# Gravitation

**Gravitation**

Introduction to Kepler's Laws

Movement is always relative. According to the energy of the moving particles, the movements are divided into two types:

« Limited movement

« Unlimited movement

In bounded motion, the total energy of the particle is negative (E < 0 ) and has one extreme point where the total energy is always equal to the potential energy of the particle, i.e. the kinetic energy of the particle becomes zero.

If the eccentricity is e ≥ 1, then E > 0 means that the body has unlimited motion. A parabolic path has eccentricity e = 1 and a hyperbolic path has eccentricity e > 1.

**Kepler's laws of planetary motion can be expressed as follows: **

**The First law of Kepler - the law of orbit**

According to Kepler's first law: "All the planets revolve around the Sun in an elliptical orbit, the Sun is at one focus". The point where the planet is closest to the Sun is called perihelion (about 17 million kilometers from the Sun), and the point where the planet is farthest from the Sun is called aphelion (152 million kilometers from the Sun). An ellipse is characterized by the fact that the sum of the distances of any planet from two foci is constant.

**Kepler's laws of planetary motion **

**Kepler's Second Law - The Law of Equal Areas**

Kepler's second law states: "A radiation vector drawn from the sun to a planet sweeps out equal areas at equal intervals of time."

Since the orbit is not a circle, the kinetic energy of the planet is not constant. It has more kinetic energy near perihelion and less kinetic energy near Apel means higher velocity at perihelion and lower velocity (v_{min}) at aphelion. If r is the distance of the planet from the Sun at perihelion (r_{min}) and aphelion (r_{max}), then

r_{min }r_{max }= 2a × (the length of the major axis of the ellipse). . . . . . . (1)

**Kepler's second law **

Kepler's second law - the plane law

Using the law of conservation of momentum, the law can be verified. The angular momentum at any time can be represented as L = mr^{2}ω.

Now consider a small area ΔA described by a short time interval Δt and subtended by an angle Δθ. Let the radius of curvature of the road be r, then

then the length of the arc covered = r Δ θ

ΔA = [r. (r. Δ θ)] = r^{2}Δθ

Therefore, $\frac{\u2206A}{\u2206t}$= $\left[\frac{1}{2}{r}^{2}\right]$$\frac{\u2206\theta}{\u2206t}$

Taking limits on both sides as, Δt→0, we get,

Now, due to the conservation of momentum, L is constant.

So $\frac{dA}{dt}$ = constant

The area swept at regular intervals is constant.

Kepler's second law can also be stated as follows: "The regional speed of a planet in an elliptical orbit around the Sun remains constant, which means that the angular momentum of the planet remains unchanged". Since momentum is constant, all planetary motions are planar motions, a direct result of centripetal force.

**Kepler's third law - The Law of Periods:**

According to Kepler's periodic law: "The Square of the time it takes a planet to revolve around the Sun in an elliptical orbit is proportional to the cube of its semi major axis".

The shorter the planet's orbit around the sun, the shorter the time it takes to make one revolution. Using the equations of the law of gravity of Newton and the laws of motion, the third law of Kepler takes a more general form.

**P ^{2 }= $\frac{4{\pi}^{2}}{\left[G\left({M}_{1}+{M}_{2}\right)\right]}\times {a}^{3}$**

Where M_{1} and M_{2} are the masses of the two orbiting objects in solar masses.

**Newton's law of universal gravitation **

Sir Isaac Newton proposed the universal law of gravitation in 1687 and used it to explain the observed motion of the planets and moons. This article introduces Newton's law of universal gravitation.

According to Newton's law of universal gravitation, every particle in the universe attracts every other particle with a force directly proportional to the product of the masses and inversely proportional to the square of the distance between them.

The universal gravitation equation thus takes the form

**Universal Gravitational Equation **

Newton's conclusion about the magnitude of the gravitational force is symbolically summarized as follows

The ratio constant (G) in the above equation is called the universal gravitational constant. Henry Cavendish determined the exact value of G experimentally. The value of G is found to be G = 6.673 x 10^{-11} N m^{2}/kg^{2}.

The Universal Law of Gravitation can explain almost everything, from how an apple falls from a tree to why the Moon orbits the Earth. Watch the video and understand the beauty of the Universal Law of Gravitation.

**Universal Law of Gravitation **

Gravitational constant

It is very difficult to accurately measure the value of the gravitational constant. Henry Cavendish developed a clever device to measure the gravitational constant.

Gravitational constant:

As shown in the figure, masses m and m' are attached to each end of the beam. The beam is attached to a strong support with a rope. The thread is tied in the middle of the beam so that it reaches balance. Now two large masses M' and M are placed next to them. The gravitational force between the two pairs of masses causes the wire to twist so that the amount of twist is balanced by the gravitational force. Gravitational force can be measured with proper calibration. Since we know the value of the masses and the distances between them, the only unknown quantity in the law of universal gravitation is G. Thus, the value of G is calculated from the measured quantities.

An example of a universal gravity solution

Calculate the force of attraction between the Earth and a 70 kg person standing at sea level 6.38 x 10^{6 }m from the center of the Earth. Solution:

Considering:

m_{1} is the mass of the Earth, which is 5.98 x 10^{2} kg

m_{2} is the mass of a person who is 70 kg

d = 6.38 x 10^{6 }m

G-value = 6.673 x 10^{-11} N m^{2}/kg^{2}

Substituting the values in the gravitational force formula, we get

**Weight and the Gravitational Force:**

In Newton's law of gravity, we noticed that mass is the deciding factor. We think of mass and weight as the same, but they are actually different. Weight is the gravitational force acting on an object of a certain mass. The weight of an object is obtained by multiplying the mass of the object m by the acceleration g due to the earth's gravity. The gravitational acceleration measured at the Earth's surface is found to be approximately 980 cm/s/s. A measure of how much material is in an object is called mass, while weight is a measure of the gravitational force acting on the material in a gravitational field. Therefore, mass and weight are proportional to the acceleration of the other like a proportionality constant of gravity. Therefore, it is observed that the mass of a particular object is constant, but the weight depends on the location of the object. To better understand, let's consider the following example, suppose we transported a body with mass m to the surface of Neptune, the gravitational acceleration would change because the radius and mass of Neptune are different from those of Earth. So our object has a mass m on the surface of both Earth and Neptune, but it weighs much more on the surface of Neptune because the gravitational acceleration there is 11.15 m/s^{2}.

**Universality of Gravity:**

Gravitational interaction exists not only between the Earth and other objects, but also exists between all objects with intensity directly proportional to the product of their masses. The law of universal gravitation helps scientists studies the orbits of the planets. Small disturbances in the elliptical motion of the planet are easily explained by the fact that all objects interact with each other under the influence of gravity.

· Why doesn't the moon hit the earth?

Velocity and gravitational forces keep the Moon in a constant orbit around the Earth. The moon seems to float in the sky, unaffected by gravity. However, the reason why the Moon remains in orbit is precisely because of gravity. In this video, you can clearly understand why the moon does not fall to the earth.

· Is gravity the same everywhere on Earth?

· Gravity is not the same everywhere on earth. In areas with greater underground mass, gravity is slightly stronger than in areas with less mass. NASA uses two spacecraft to measure changes in Earth's gravity. These spacecraft are part of the Gravity Recovery and Climate Experiment (GRACE) mission.

**Gravitational variations on Earth **

Blue areas have weaker gravity, while red areas have slightly stronger gravity.

**The area in blue has weaker gravity while the area in red has slightly stronger gravity.**

**Acceleration Due to Gravity:**

**Gravity:**

Gravity is the force by which the Earth pulls an object towards its center. Consider two mass bodies, ma, and mb. If equal forces are applied to two objects, the force corresponding to the mass is given by the formula

m_{b }= m_{a [$\frac{{a}_{A}}{{a}_{B}}$}_{]}

This is called the inertial mass of the body.

Due to the gravity of two bodies

The upper mass is called the gravitational mass of the body. According to the equivalence principle, inertial mass and gravitational mass are identical. We use this to derive the acceleration due to gravity shown below.

Suppose a body [test mass (m)] is dropped from a height 'h' above the ground [source mass (M)]; As it approaches the ground, it begins to move downward with increasing speed.

We know that the speed of an object is only changed by a force; in this case, the force is due to gravity.

Under the influence of gravity, the body begins to accelerate towards the center of the Earth, which is at a distance "r" from the test mass. Then ma = GMm/r2 (applying the equivalence principle)

The above acceleration is due to the pull of the earth, so we call it the acceleration due to gravity; it does not depend on the mass of the test. Its value near the ground is 9.8 ms-².

Therefore, the acceleration due to gravity (g) is given by = GM/r². The formula for acceleration due to gravity.

The force acting on the body under the influence of gravity is given by the formula f = mg

Where f is the force acting on the body, g is the acceleration due to gravity and m is the mass of the body.

According to the law of universal gravitation

f = GmM/(r h)²

where,

f = force between two bodies

G = universal gravitational constant (6.67 × 10^{-11} Nm²/kg²)

m = mass of the object

M = Earth's mass

r = radius of the Earth

h = height at which the object is above the ground

Since the height (h) is negligible compared to the radius of the earth, let's reformulate the equation as follows:

f = GmM/r²

Now that both expressions are equated,

mg = GmM/r²

⇒ g = GM/r²

Therefore, the formula for the acceleration due to gravity is g = GM/r²

Note: This depends on the mass and radius of the earth.

This helps us understand the following.

All bodies experience the same acceleration due to gravity, regardless of their mass.

Its value on earth depends on the mass of the earth, not the mass of the object.

**Acceleration Due to Gravity on the Surface of Earth**

Earth is assumed to be a uniform solid sphere with a mean density. We know that,

Density = mass/volume

Then, ρ = M/[4/3 πR^{3}]

⇒ M = ρ × [4/3 πR^{3}]

We know that g = GM/R^{2}.

On substituting the values of M, we get,

g = 4/3 [πρRG]

At any distance ‘r’ from the centre of the earth,

g = 4/3 [πρRG]

The value of acceleration due to gravity ‘g’ is affected by

« Altitude above the earth’s surface.

« Depth below the earth’s surface.

« The shape of the earth.

« Rotational motion of the earth.

**Variation of g with Height:**

Acceleration due to gravity at a height (h) from the surface of the earth

Consider a test mass (m) at a height (h) from the surface of the earth. Now, the force acting on the test mass due to gravity is

F = GMm/(R+h)^{2}

Where M is the mass of the earth, and R is the radius of the earth. The acceleration due to gravity at a certain height is ‘h’, then

mgh= GMm/(R+h)^{2}

⇒ gh= GM/[R^{2}(1+ h/R)^{2} ] . . . . . . (2)

The acceleration due to gravity on the surface of the earth is given by

g = GM/R^{2 }. . . . . . . . . (3)

On dividing equations (3) and (2), we get

gh = g (1+h/R)^{-2}. . . . . . (4)

This is the acceleration due to gravity at a height above the surface of the earth. Observing the above formula, we can say that the value of g decreases with an increase in the height of an object, and the value of g becomes zero at an infinite distance from the earth.

**Approximation Formula:**

From equation (4)

when h << R,

the value of g at height ‘h’ is given by gh = g/(1 – 2h/R)

**Variation of g with Depth:**

Variation of Acceleration due to Gravity with Depth

Consider a test mass (m) taken to a distance (d) below the earth’s surface, the acceleration due to gravity at that point (gd) is obtained by taking the value of g in terms of density.

On the surface of the earth, the value of g is given by

g = 4/3 × πρRG

At a distance (d) below the earth’s surface, the acceleration due to gravity is given by

gd = 4/3 × πρ × (R – d) G

On dividing the above equations, we get,

gd = g (R – d)/R

When the depth d = 0, the value of g on the surface of the earth gd = g.

When the depth d = R, the value of g at the centre of the earth gd = 0.

**Variation of g Due to the Shape of the Earth**

As the earth is an oblate spheroid, its radius near the equator is more than its radius near the poles. Since for a source mass, the acceleration due to gravity is inversely proportional to the square of the radius of the earth, it varies with latitude due to the shape of the earth.

From the above equation, it is clear that acceleration due to gravity is more at the poles and less at the equator. So if a person moves from the equator to the poles, their weight decreases as the value of g decreases.

**Variation of g Due to Rotation of Earth**

Consider a test mass (m) is on a latitude making an angle with the equator. As we have studied, when a body is under rotation, every particle in the body makes circular motions about the axis of rotation. In the present case, the earth is under rotation with a constant angular velocity ω, then the test mass moves in a circular path of radius ‘r’ with an angular velocity ω.

^{2}). Gravity is acting on the test mass towards the centre of the earth (mg).

We know from the parallelogram law of vectors if two co-planar vectors are forming two sides of a parallelogram, then the resultant of those two vectors will always be along the diagonal of the parallelogram.

Applying the parallelogram law of vectors, we get the magnitude of the apparent value of the gravitational force at the latitude.

(mg′)^{2} = (mg)^{2} + (mrω^{2})^{2} + 2(mg) (mrω^{2}) cos(180 – θ) . . . . . . (1)

We know ‘r’ is the radius of the circular path and ‘R’ is the radius of the earth, then r = Rcosθ.

Substituting r = R cosθ. we get,

g′ = g – Rω^{2}cos2θ

Where g′ is the apparent value of acceleration due to gravity at the latitude due to the rotation of the earth, and g is the true value of gravity at the latitude without considering the rotation of the earth.

At poles, θ = 90°⇒ g’ = g.

At the equator, θ = 0° ⇒ g′= g – Rω^{2}.

**Important Conclusions on Acceleration Due to Gravity**** **

· For an object placed at a height of h, the acceleration due to gravity is less as compared to that placed on the surface.

· As depth increases, the value of acceleration due to gravity (g) falls.

· The value of g is more at the poles and less at the equator.

**Gravitational field: **

In the case of non-contact force, the source mass and the test mass interact via a gravitational field. You can think of the gravitational force as a "command" and the gravitational field as a dialogue or speech used to give an order.

**Gravitational field intensity **

The strength of the gravitational field is called the gravitational field strength. This is the gravitational force acting on the unit mass.

where r cap represents the position vector of the test mass from the source mass. The intensity of the gravitational field depends only on the mass of the source and the distance of a unit of the test mass from the mass of the source.

The unit of gravitational field strength is N/kg.

The scale gives [M^{0}L¹T–²]. The scale of gravitational field strength is identical to that of acceleration (from a gravitational perspective, we prefer to call it acceleration due to gravity).

The principle of superposition extends to the intensity of the gravitational field, for example

E = E_{1} + E_{2} + E_{3 }+ . . . . . . . + E_{n}

Thus, E_{1}, E_{2}, E_{3}, . . . . . In are the gravitational field intensities at the point due to the nth particle of the system.

In the system, mass is always distributed in two different ways:

- Discrete mass distribution
- Continuous mass distribution

dE is the intensity of the gravitational field due to the main mass dm. The gravitational field formula is expressed as:

g = F/m

where, F = gravitational force and m = mass of the object.

Gravitational Field Intensity of a Point Mass:

Consider a point mass M, the gravitational force at a distance 'r' from it is given by the formula

Gravitational Intensity of the Gravitational Field created by the Ring:

Consider a ring of mass M with radius 'a'; the gravitational field at a distance x along its axis is found as:

**Gravitational field intensity **

Consider a small element along the circumference of the ring with mass 'dm'; the field strength resulting from that length element is given by the formula,

dE = Gdm/r²

The vertical components of the fields cancel each other due to ring symmetry, and only the horizontal components are preserved and added.

**Gravitational field due to Uniform Spherical Shell:**

Consider a thin uniform spherical shell in space of radius 'R' and mass 'M'. A 3D object divides space into three parts:

- Inside the spherical shell.
- On the surface of the spherical shell.
- Outside the spherical shell.

Our problem is to find out the value of gravitational field intensity in all these 3 regions

**Outside the spherical shell **

Consider a unit test mass at a point "P" at a distance "r" from the center of the spherical shell. Draw an imaginary spherical shell with point 'P' on its surface.

As we know, the gravitational field intensity at a point depends only on the mass of the source and the distance of the point from the mass of the source. We can say that the initial mass in the imaginary sphere is M and the separation distance is 'r'. That's what we get

E = -GM/r²

⇒ E ∝ -1/r²

**On the surface of the spherical shell **

Consider a unit test mass at a point "P" on the surface of the spherical shell at a distance "r" from the center of the spherical shell, where r = R. As mentioned above, the gravitational field intensity at the surface of the spherical shell is given by spherical shell

E = -GM/R²

**Inside a spherical shell **

If you consider a point inside a spherical shell, the entire mass of the shell is above the point. Draw an imaginary spherical shell around point 'P'; to leave this imaginary realm is zero.

We know that if the mass of the source is zero, the intensity of the gravitational field is also zero.

∴ E = 0

**Gravitational field due to a Uniform Solid Sphere:**

Consider a uniform solid sphere of radius "R" and mass "M". Let's find the value of the gravitational field strength in all three regions:

- Within a solid sphere.
- On the surface of a solid sphere.
- Off a certain ball.

To find the gravitational field intensity at a point "P" at a distance "r" from the center outside the solid sphere, consider an imaginary sphere around P enclosing the entire mass "M".

∴ E = – GM/r²

⇒ E ∝ -1/r²

**On the surface of a solid sphere **

To find the gravitational field intensity at point "P" on the surface of a solid sphere,

Then E = -GM/R² ⇒ E = Constant.

**Within a solid sphere **

Find the gravitational effect at a point 'P' inside a uniform solid sphere at a distance 'r' from the center of the sphere. If we draw an imaginary sphere around this point, the mass of that imaginary sphere is given by 'm'.

For space (4/3) πR³, the existing mass is M; for volume (4/3) πr³ the existing mass is "m".

Since the density of a solid sphere remains the same at all times,

m = M × (r³/R³)

Then the gravitational field intensity at point "P" inside the solid sphere at a distance "r" from the center of the sphere is given by:

E = -Gm/r²

Where m is the initial mass on an imaginary sphere drawn around point "P". Substituting the value of m in the above equation, we get

E = -GMr/R³

⇒ E ∝ -r

**Gravitational potential energy - formulas, derivatives, solved problems **

Gravitational potential energy is the energy that an object possesses or gains from its change in position when it is in a gravitational field. Simply put, gravitational potential energy is the gravitational force or energy associated with gravity.

The most common example to help you understand the concept of gravitational potential energy is that you take two pencils, one of which is placed on a table and the other is held above the table. Now we can argue that a pencil in the air has more gravitational potential than a pencil on the table

**Gravitational Potential Energy:**

If a body of mass (m) is moved from infinity to a point under the gravitational influence of the initial mass (M) without accelerating it, the amount of work done in moving it to the initial field is stored in the form of potential energy. . . . This is called gravitational potential energy. It is denoted by the symbol Ug.

Explanation:

We know that the potential energy of a body at a particular location is defined as the energy stored in the body at that location. When the position of the body changes under the influence of external forces, the change in potential energy is equal to the amount of work done by the forces acting on the body. The work done by gravity does not depend on the path of change of position, so the force is a conservative force. All such forces also have a certain potential.

The effect of gravity on a body at infinity is zero; therefore, the potential energy is zero, called the reference point.

Gravitational Potential Energy Formula

The equation for gravitational potential energy is:

⇒ G_{PE} = mgh

Where,

- m is the mass in kilograms
- g is the acceleration due to gravity (9.8 on earth)
- h is the height above the ground in metres

**Derivation of Gravitational Potential Energy Equation:**

Consider a source mass ‘M’ is placed at a point along the x-axis; initially, a test mass ‘m’ is at infinity. A small amount of work done in bringing it without acceleration through a very small distance (dx) is given by

dw = Fdx

Here, F is an attractive force, and the displacement is towards the negative x-axis direction, so F and dx are in the same direction. Then,

dw = (GMm/x2)dx

Integrating on both sides

If r_{i}> r_{f} then ΔU is negative.

Since the work done is stored as its potential energy U, the gravitational potential energy at a point which is at a distance ‘r’ from the source mass is given by;

U = -GMm/r

If a test mass moves from a point inside the gravitational field to another point inside the same gravitational field of source mass, then the change in potential energy of the test mass is given by;

ΔU = GMm (1/r_{i }– 1/r_{f})

If r_{i}> r_{f} then ΔU is negative.

**Expression for Gravitational Potential Energy at Height (h) –**

Derive ΔU = mgh.

If a body is taken from the surface of the earth to a point at a height ‘h’ above the surface of the earth, then r_{i} = R and r_{f}= R + h, then,

ΔU = GMm [1/R – 1/(R+h)]

ΔU = GMmh/R(R + h)

When h<<R, then R + h = R and g = GM/R^{2}.

On substituting this in the above equation, we get,

Gravitational Potential Energy ΔU = mgh

Note:

The weight of a body in the center of the earth is zero because the value of g in the center of the earth is zero.

At a point in the gravitational field where the gravitational potential energy is zero, the gravitational field is zero.

What Is Gravitational Potential?

The amount of work done in moving a unit test mass from infinity into the gravitational influence of source mass is known as gravitational potential.

Simply, it is the gravitational potential energy possessed by a unit test mass.

⇒ V = U/m

⇒ V = -GM/r

⇒ Important Points:

The gravitational potential at a point is always negative, and V is maximum at infinity.The SI unit of gravitational potential is J/s

Relation between Gravitational Field Intensity and Gravitational Potential

**Integral Form:**

V = - $\int \underset{E}{\to}.\underset{dr}{\to}$

( If E is given and V has to be found using this formula)

**Differential Form**:

E = -dV/dr (If V is given and E has to be found using this formula)

(components along x, y, and z directions).

**Gravitational Potential of a Point Mass**

Consider a point mass M, the gravitational potential at a distance ‘r’ from it is given by;

V = – GM/r.

Gravitational Potential of a Spherical Shell

Consider a thin uniform spherical shell of the radius (R) and mass (M) situated in space. Now,

- Case 1: If point ‘P’ lies inside the spherical shell (r<R):

As E = 0, V is a constant.

The value of gravitational potential is given by, V = -GM/R.

Case 2: If point ‘P’ lies on the surface of the spherical shell (r=R):

On the surface of the earth, E = -GM/R^{2}.

Using the relation

- Case 1: If point ‘P’ lies inside the spherical shell (r<R):

As E = 0, V is a constant.

The value of gravitational potential is given by, V = -GM/R.

- Case 2: If point ‘P’ lies on the surface of the spherical shell (r=R):

On the surface of the earth, E = -GM/R^{2}.

Using the relation

over a limit of (0 to R), we get,

Gravitational Potential (V) = -GM/R.

- Case 3: If point ‘P’ lies outside the spherical shell (r>R):

Outside the spherical shell, E = -GM/r^{2}.

Using the relation

V = - $\int \underset{E}{\to}.\underset{dr}{\to}$

over a limit of (0 to R), we get,

Gravitational Potential (V) = -GM/R.

Case 3: If point ‘P’ lies outside the spherical shell (r>R):

Outside the spherical shell, E = -GM/r^{2}.

Using the relation

V = - $\int \underset{E}{\to}.\underset{dr}{\to}$

**Gravitational Potential of a Uniform Solid Sphere:**

Consider a thin, uniform solid sphere of radius (R) and mass (M) situated in space. Now,

- Case 1: If point ‘P’ lies inside the uniform solid sphere (r < R):

Inside the uniform solid sphere, E = -GMr/R^{3}.

Using the relation

V = - $\int \underset{E}{\to}.\underset{dr}{\to}$

over a limit of (0 to r).

The value of gravitational potential is given by,

V = -GM [(3R^{2} – r^{2})/2R^{2}]

- Case 2: If point ‘P’ lies on the surface of the uniform solid sphere ( r = R ):

On the surface of a uniform solid sphere, E = -GM/R^{2}. Using the relation

V = - $\int \underset{E}{\to}.\underset{dr}{\to}$

over a limit of (0 to R) we get,

V = -GM/R.

- Case 3: If point ‘P’ lies outside the uniform solid sphere ( r> R):

Using the relation over a limit of (0 to r), we get, V = -GM/R.

- Case 4: Gravitational potential at the centre of the solid sphere is given by

V =(-3/2) × (GM/R).

**Gravitational Self Energy**

The gravitational self-energy of a body is defined as the work done by an external agent in assembling the body from the infinitesimal elements that are initially at an infinite distance apart.

Gravitational self-energy of a system of ‘n’ particles:

Let us consider n particle system in which particles interact with each other at an average distance ‘r’ due to their mutual gravitational attraction; there are n(n – 1)/2 such interactions, and the potential energy of the system is equal to the sum of the potential energy of all pairs of particles, i.e.,

**Solved Problems**

Example 1. Calculate the gravitational potential energy of a body of mass 10 kg and is 25 m above the ground.

Solution:

Given, Mass m = 10 Kg and Height h = 25 m

G.P.E is given as,

U = m × g × h = 10 Kg 9.8 m/s^{2} × 25 m = 2450 J.

**Escape Velocity:**

Escape velocity is the minimum velocity at which a mass would have to project from the ground to escape the earth's gravitational field. Escape velocity, also known as escape velocity, is defined as:

The minimum speed required for an object to break free from the gravitational pull of a massive object.

For example, if we think of the Earth as a massive body. Escape velocity is the minimum speed an object must reach to overcome the Earth's gravitational field and fly to infinity without falling back. It purely depends on the distance of the object from the massive body and the mass of the massive body. The greater its mass, the greater the distance, the greater the escape velocity.

For all massive bodies, such as planets and stars, which are spherically symmetric in nature, the escape velocity at any distance is mathematically expressed as:

where,

- woe is the escape velocity
- G is the universal gravitational constant.
- M is the mass of the massive body (the body from which the object should escape)
- r is the distance from the center of the massive body to the object

It may be noted here that the above relation does not depend on the mass of the object escaping from the massive body.

**Derivation of escape velocity **

In general escape, velocity is achieved when an object is moving at a velocity at which the arithmetic sum of the object's gravitational potential energy and its kinetic energy is zero. This means that to reach infinity, the object would have to have more kinetic energy than gravitational potential energy. The easiest way to derive the formula is to use the concept of energy conservation. Suppose an object tries to escape a planet (which is uniformly circular in nature) by moving away from it. The main force acting on such an object is the planet's gravity. As we know, kinetic energy (K) and gravitational potential energy (Ug) are the only two types of energy. Using the principle of energy conservation, we can write:

By the principle of conservation of energy, we can write:

(K+Ug)i=(K+Ug)f

Where,

K=1/2mv^{2}

Here U_{gf} is zero as the distance is infinity and K_{f} will also be zero as the final velocity will be zero. Thus, we get:

The minimum velocity required to escape from the gravitational influence of a massive body is given by:

The escape speed of the earth at the surface is approximately 11.186 km/s. That means “an object should have a minimum of 11.186 km/s initial velocity to escape from earth’s gravity and fly to infinite space.”

Ideally, if you can jump at an initial speed of 11,186 km/s, you can orbit in space!

**Unit of escape velocity **

The unit of escape velocity, or escape velocity, is meter per second (m.s-¹), which is also the SI unit of escape velocity.

**Earth Satellites:**

Satellite: An object that orbits the sun, the earth, or some other massive body is called a satellite. Satellites are divided into two main types, one is natural and the other is man-made. Some examples of natural satellites are planets, moons and comets. Jupiter has 67 natural satellites. The Earth has one permanent natural satellite, the Moon we know, which causes tides in the ocean. Sometimes other objects (such as asteroids) can move into temporary earth orbits and become natural satellites within a certain distance. Apart from these, there are many artificial satellites placed in orbit on the Earth, which are used for various applications of communication and data collection. As the term itself suggests, an artificial satellite is one that is placed in space by humans and follows the orbit of natural satellites. Because they have a very large field of view, they can collect data much faster than ground-based instruments. In addition, clouds, dust and other obscurations do not obstruct their view of space beyond the Earth, which allows a satellite to observe space much more effectively than with telescopes on Earth. Satellite . More than 2,500 man-made satellites currently orbit the Earth. Most of them are of Russian origin. You may be wondering why none of these satellites collide with each other because of the volume. In fact, it is quite possible. Although care is taken to guide the satellite into specific orbits so that collisions never occur, these orbits can vary in nature. Many international organizations prevent such events. However, in 2009, a pair of Russian and American satellites collided for the first time!

Satellites are launched for a specific purpose related to multiple uses such as communications, research, weather forecasting, and intelligence. When launched into space, all different types of satellites follow the same principles of physics and are governed by the same mathematical equations. There are two types of artificial satellites based on their purpose. These are geostationary satellites and polar satellites.

**Types of satellites: **

**Geostationary satellite: **

· These satellites are placed in an orbit around 35,800 kilometers from the Earth. They rotate in the same direction as the Earth, and one revolution of such satellites equals one day on Earth (about 2 hours). This means that when viewed from Earth, these satellites appear to be in the same place all the time. Hence the name "geostationary" satellites. These satellites are used as communication satellites and in weather-based applications. Polar Satellite:

« Unlike geostationary satellites, polar satellites orbit the Earth in a north-south direction. They are very useful in applications that require an outdoor view of the entire globe in one day. Since the whole globe moves under them, it is easy to do. They are used in

Weather applications where weather and climate-based disasters can be predicted at short notice. They are also used as relay stations.

**Polar satellite **

The International Space Station (ISS) was launched in 1998. It is a habitable artificial satellite that can sometimes be seen with a clear sky. It serves as a laboratory, observatory, and landing base for potential expeditions.

**The Nature of the Satellite Projectile: **

The most important thing to understand about satellites is that they are, after all, munitions. Any object that is only affected by gravity is called a satellite. Gravity is the only thing that affects a satellite after it is launched into orbit. Movement of satellites

To understand this concept clearly, let's use the example of sending a satellite from the top of Mount Newton, a hypothetical location far above the effects of air resistance. Newton was the first scientist who present the concept that if an object is launched with sufficient speed, it will begin to orbit the Earth. This object would experience a gravitational force that would try to pull it down moving horizontally tangent to the ground. If the launch velocity is less than the escape velocity, it will fall back to the ground.

If the projectile is fired at full speed with escape velocity, it will fall into orbit outside the earth and begin to circle the earth; dashed line C represents such an object. If the object is launched at a higher speed, it will still orbit the Earth, but will now have an elliptical orbit; dashed line D represents such an object. It may also be possible for an object to be shot at such a speed that it escapes the gravitational pull of the earth and becomes a free body; the solid line E represents such an object. Objects C and D never fall back to the ground even though they are constantly drawn to it because our earth is a round body.

All of this observation raises the very fundamental question of how much velocity is required to shoot an object out of Earth's lower atmosphere and deposit it outside, still in the gravitational field. We get the answer by observing the main aspect of the Earth, and measuring its curvature. It has been measured that for every 8,000 meters traveled along the earth's horizon, the surface drops about 5 meters. So, applying basic math, we can assume that if a bullet wants to go around the sun, it must have.

**Energy from an Orbiting Satellite: **

Satellites revolve around a massive central body in either circular or elliptical orbits. A satellite orbiting the Earth travels at a constant speed and fixed height in its orbit, moving at a tangential speed that allows it to fall at the same rate as the Earth's rotation. Gravitational force acts perpendicular to the direction of movement of the satellite throughout the entire trajectory.

According to the work-energy theorem, the total initial mechanical energy of a system plus the work did by some external force equals the final mechanical energy.

For satellites, gravity is the only external force, and since gravity is considered a conservative force, the Wext term is zero. The equation can be simplified as follows:

In other words, the sum of the kinetic and potential energy of a system is constant, while the energy varies between kinetic and potential energy.

**Analysis of a Circular Orbit: **

During the rotational movement around the Earth, the satellite remains at a certain distance from the Earth's surface all the time. Since the tangential velocity is a function of the radius of the orbit, the velocity remains constant, as does the kinetic energy. Also, since the potential energy depends on the height of the object, which remains constant in this case, the potential energy therefore remains constant at all times. Thus, the total mechanical energy, or KE PE, remains constant. Energy from an orbiting satellite

**Analysis of Elliptical Orbits **

The total mechanical energy of a satellite in elliptical motion also remains constant, as in circular motion, but unlike circular motion, the energy of the satellite in elliptical motion changes form. It is known that the tangential velocity of a body orbiting the Earth is inversely proportional to the square root of its orbital radius, and the kinetic energy also decreases as the radius increases and is inversely proportional to the orbital radius. Therefore, the potential energy increases as the height of the object increases and thus increases as the radius of the orbit.

**Energy from an orbiting satellite **

The movement of the satellite around the Earth is considered circular. In this section, we derive the expression for kinetic energy, potential energy, and total mechanical energy along a circular path around the Earth. Energy from an orbiting satellite

The tangential velocity of a satellite orbiting the Earth can be given as follows

**Energy from an Orbiting Satellite: **

Where M is the mass of the earth, R is the radius of the earth, and h is the height above the earth where the object is located.

Thus, the kinetic energy (mass m) of the satellite in a circular orbit at speed v can be written as follows

As per our assumption, the gravitational potential energy at infinity is considered to be zero, so, the potential energy at distance (R +h) from the center of the earth can be written as

The kinetic energy here is positive whereas the potential energy is negative. However, in magnitude, the kinetic energy is half the potential energy, so the total energy E is

The total energy of a circularly orbiting satellite is thus negative but twice is the magnitude of the positive kinetic energy.

**Consideration **

Weightlessness is a term used to describe the feeling of complete or near complete weightlessness. Astronauts orbiting the Earth often experience the sensation of weightlessness. The feelings experienced by astronauts in orbit are the same as those experienced by anyone temporarily above a seat on an amusement park ride. The reasons for the feeling of weightlessness are the same in both cases.

**Why do we feel burdened? **

Weight is the feeling that a person experiences when his body does not touch external objects. In other words, the sensation of weightlessness occurs when all contact forces are removed. These feelings are common in free fall mode.

During free fall, the only force acting on the body is gravity. Since gravity is a non-contact force, it cannot be felt without a counter force. That's why you feel weightless when free falling.

It is important to remember that being weightless is just a feeling, not a reality that corresponds to a weightless person. Balance has little to do with weight and a lot to do with the presence and absence of contact forces.

The Elevator Test by Otis L. Evaderz

Did you know that a scale doesn't measure weight? Rather, the scale reading is a measure of the upward force exerted by the scale to balance the downward force of gravity on the person standing on the scale. When the body is in equilibrium, these two forces are in balance. The upward force on the person is equal to the downward force of gravity. In such cases, the weight reading corresponds to the person's weight. However, if the person standing on the scale jumps up and down, the scale reading changes rapidly. When bouncing, the body accelerates. As a result, the upward force of the scale changes. Does this mean the weight will also change? Of course not! You weigh the same. Only the weight reading changes, because it does not measure your weight, but the contact force acting on the body. Otis L. Evaderz performed the famous elevator experiment by riding up and down an elevator while standing on a bathroom scale. He noticed that the reading on the scale was different when he accelerated up and down and when he was at rest or moving at a constant speed. We know that the scale reading is a measure of the upward normal force, so its value can be predicted at different stages of the movement. The value of the normal force acting on Otis' 80 kg body could be predicted if the acceleration is known. This prediction is made using Newton's second law of motion.

Consideration

In the diagram, the 80 kg Otis is moving at constant speed (A), accelerates up (B), accelerates down (C), and falls freely (D) after the elevator cable breaks. The normal force is greater than gravity because the elevator is accelerating upward (B). And it is less than gravity with downward acceleration (C and D) and equal to gravity without acceleration.

Since normal is responsible for detecting weight, the lifter feels his normal weight in case A, slightly heavier in case B, and less than normal weight in case C. For D, the rider felt a weight, without an external force he would not feel the weight. Finally, it can be stated that the rider weighs the same in all four cases, but the sensation of heaviness he feels is different. The weight of the rider varies during the ride.

Why do astronauts feel weighed down in space? Astronauts orbiting in space feel weightless because there is no external contact force in space to push or pull their bodies. Gravity is the only force acting on their bodies. Gravity, which is force acting at a distance, cannot be felt and therefore does not give a sense of weight.

Do astronauts experience weightlessness because there is no gravity in space? Many students assume that astronauts feel weightless because there is no gravity in space. This is not true. If this were true, it would violate the principles of circular motion. If someone believes that the lack of gravity in space is the reason for weightlessness, should they come up with a reason for astronauts orbiting in space?

Is gravity in space less than gravity on Earth? The gravity acting on an astronaut in space is certainly less than the gravity on the Earth's surface. But it is not small enough to account for the drastic weight. Consider a space station orbiting about

00 km above the earth when the value of g at that location decreases from 9.8 m/s2 to about 8.7 m/s2. While this certainly reduces weight, it does not account for the absolute weightlessness experienced by astronauts. Their sense of absolute weightlessness comes from the fact that they have no surface to support them as they freely fall to Earth.