Why is gravitational potential energy typically represented as a negative value?

It is always negative because gravitational force is always attractive in nature.

What are two key factors that influence whether a planet can sustain an atmosphere?

Surface Temperature of the Planet Acceleration Due to gravity at the surface of the planet

Why does a body experience weightlessness at the center of the Earth?

At the centre of the Earth g=0 W=mg=0

Write characteristics of Gravitational force?

1.It is a central force 2.It is a conservative Force. 3.It obeys inverse square law. 4.It is always attractive in nature.

What fundamental law of physics does Kepler's second law represent?

Law of conservation of angular momentum

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Gravitation: Download Important IIT JEE & NEET Questions

Gravitation is a fundamental force of nature that governs the motion of objects throughout the universe. It is the attractive force that draws two bodies toward one another, resulting in a pull that depends on their masses and the distance separating them. This force not only keeps us anchored to the Earth but also plays a vital role in the movement of celestial bodies, such as planets, moons, and stars.

The understanding of gravitation has significantly evolved over the centuries. In the late 17th century, Sir Isaac Newton was among the first to mathematically describe this force through his Universal Law of Gravitation.

1.0Newton Law of Gravitation

Gravitation is the property of matter due to which two bodies always attract each other.
It states that every particle in the universe attracts all other particles with a force which is directly commensurate to the product of their masses and is inversely commensurate to the square of the distance between them.
This formula is given for point masses. For spherical objects (planets), the distance between centres is taken.

$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$

G=universal Gravitational Constant,Its value is the same throughout the universe; G independent of the nature and size of the bodies; it does not depend even upon the nature of the medium between the bodies.
Value of G in $SI = 6.67 \times 1 0^{- 11} Nm^{2} kg^{- 2} or CG.S = 6.67 \times 1 0^{- 8} dyne cm^{2} g^{- 2}$
Dimension of $G = [M^{- 1} L^{3} T^{- 2}]$
Forces due to Multiple Particles is given as, $F_{n e t} = F_{1} + F_{2} + F_{3} + \dots\dots$

Law of Gravitation in Vector Form

$F_{12} = - G \frac{m _{1} m _{2}}{r ^{2}} \overset{r}{^}$

Related Video:

2.0Gravitational Field

It is the region around a mass where its gravitational effects can be observed. The gravitational field due to a body extends till infinity.
Strength of the gravitational field decreases on moving away from a body.

(i) In an infinite gravitational field, the acceleration is zero.

(ii) Gravitational field ∝ mass

Gravitational field is conservative in nature.
There are two ways to measure the strength of the gravitational field.

(i) Gravitational field intensity.

(ii) Gravitational potential

Gravitational Field Intensity

It is defined as gravitational force exerted on a unit mass positioned in a gravitational field.
It is a vector quantity.Its direction is same as that of gravitational force.
$I = \frac{F}{m} \Rightarrow Unit - N / K g$
Dimensional Formula $- [M^{0} L^{1} T^{- 2}]$
Gravitational Field Intensity owed to a point mass,is given as,

$I = \frac{GM}{r ^{2}} (- \overset{r}{^})$

Negative sign indicates that the force due to the gravitational field is attractive in nature.
Gravitational Field Intensity due to multiple particles,

$I_{n e t} = I_{1} + I_{2} + I_{3} + \dots\dots\dots\dots\dots$

A. Solid Sphere

Outside the Sphere (r>R)

$I_{out} = \frac{GM}{r ^{2}} (- \overset{r}{^})$

At the Surface (r=R)

Gravitational field intensity at the surface of sphere

$I_{surface} = \frac{GM}{R ^{2}} (- r)$

Inside the Sphere (r<R)

Gravitational Field Intensity Inside the Sphere

$I_{in} = \frac{GM r}{r ^{3}} (- \overset{r}{^})$

Graph between $I$ and r for a solid sphere

Graph between gravitational field intensity and r

B. Hollow Sphere

Outside the Shell (r>R)

Gravitational Field Intensity Outside the Shell

$I_{out} = \frac{GM}{r ^{2}} (- \overset{r}{^})$

At the Surface (r=R)

Gravitational field intensity at the surface

$I_{surface} = \frac{GM}{R ^{2}} (- \overset{r}{^})$

Inside the Shell (r<R) I=0

Gravitational field intensity in the inside of the shell

Graph between $I$ and r for Hollow sphere

Graph between gravitational field intensity and r for a hollow sphere

3.0Acceleration Due to Gravity

If a body is placed inside a gravitational field having intensity I, gravitational force applied on it is, F=m I
Earth also pulls every object towards its center with force F=mg

On comparing both the equations we get $g = ∣ I ∣$

Where g = acceleration due to gravity I is intensity of gravitational field due to Earth.

Assuming Earth to be a Solid Sphere

$g_{s} = \frac{GM}{R _{e}^{2}} \Rightarrow 9.81 m / s^{2}$ , Same formula can be used to find acceleration due to gravity at the surface of any planet.

In terms of density $(ρ) : g = \frac{GM}{R ^{2}}$

$mass = density \times Volume$

$M = ρ \times \frac{4}{3} π R^{3}$

$g = \frac{G ( \frac{4}{3} π R ^{3} )}{R ^{2}} = \frac{4}{3} π GRρ$

Variation in acceleration due to gravity with height

Variation in acceleration due to gravity with height

$g_{h} = g [1 - \frac{2 h}{R _{e}}]$ , This formula is valid if h is upto 5% of earth's radius (320 km from earth's surface)

If h is greater than 5% earth's radius we use, $g_{h} = \frac{G M _{e}}{( R _{e} + h ) ^{2}}$

Variation in acceleration due to gravity with depth

$g_{d} = g_{s} (1 - \frac{d}{R})$ , For any depth below Earth's Surface

Note:

Valid for small or large depth.
Fractional change in value of $g \to \frac{Δ g}{g} = \frac{d}{R}$
At any depth, $g_{d} < g_{s}$
Another form that can be used $g_{d} = \frac{GM}{R ^{3}} r \to g_{d} = g_{s} \frac{r}{R}$

Graph

For (r<R)

$g = \frac{GM}{R ^{3}} r \to g = g_{s} \frac{r}{R}$

For $(r \geq R)$

$g = \frac{GM}{r ^{2}} \to g = g_{s} \frac{R ^{2}}{r ^{2}}$

Value of g is maximum at Earth's surface

Variation in acceleration due to gravity with depth

Variation in value of 'g' due to rotation of earth

Variation in value of acceleration due to gravity due to rotation of earth

$N = M g_{e ff} = M g - M ω^{2} R_{e} Cos^{2} λ$ , As one moves towards pole, apparent weight increases

$λ : Angle of Latitude$

$For Poles λ = 9 0^{\circ}$

$For Equator λ = 0^{\circ}$

$g =$ acceleration due to gravity without considering rotation of Earth

$g_{eff} =$ acceleration due to gravity considering rotation of Earth

Variation in value of "g" due to shape of Earth

Variation in value of acceleration due to gravity due to shape of Earth

$Δ g = g_{e} - g_{p} \approx 0.02 m / s^{2}$ , Earth is not a perfect sphere

$g_{surface} = \frac{GM}{R ^{2}}$

$g_{p} > g_{eq’}$ (as surface is close to center of Earth near pole as compared to equator )

4.0Gravitational Potential

It is defined as amount of work done by external agent against gravitational pull in conveying a unit mass from infinity to that point without changing its kinetic energy

$V_{P} = \frac{W _{\infty \to p}}{m}$

Motion of unit mass is considered slowly, so that no energy is spent to change the kinetic energy of the system.
At infinity, Gravitational Potential is assumed to be ZERO.

Also Read : Difference Between Mass and Weight

Gravitational Potential Due To a Point Mass:

$V = \frac{W _{e x t}}{m} = - \frac{GM}{r}$ (Negative sign in potential indicates attractive nature of force. )

Relation between Potential and Intensity

$I = - \frac{d V}{d r}$

If V is constant in a region, I = 0
This relation is valid for all conservative fields.

Standard results to remember:

Spherical shell (M, R)

Case 1. Outside the Shell $(r > R), V_{out} = - \frac{GM}{r}$

Case 2. On the Surface $(r = R), V_{Surface} = - \frac{GM}{R}$

Case 3. Inside the Shell $(r < R), V_{I n} = - \frac{GM}{R}$ , (Potential is same everywhere and is equal to its value at the surface.)

Solid sphere(M, R)

Case 1. Outside the Sphere $(r > R), V_{out} = - \frac{GM}{r}$

Case 2. On the Surface $(r = R), V_{Surface} = - \frac{GM}{R}$

Case 3. Inside the Sphere $(r < R), V_{I n} = - \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})$ , (Potential is same everywhere and is equal to its value at the surface.)

5.0Gravitational Potential Energy

The gravitational potential energy of a particle situated at a point in some gravitational field is defined as the amount of work required to bring it from infinity to that point without changing its kinetic energy.

$U = - G \frac{m _{1} m _{2}}{r}$ , negative sign shows the boundedness of the two bodies

It is a scalar quantity.
It's SI unit is joule and Dimensions are [M¹L²T^–2]
The gravitational potential energy of a particle of mass 'm' placed on the surface of earth of mass 'M' , radius 'R' is given by : $U = - G \frac{m _{1} m _{2}}{R}$

Gravitational Potential Energy for Three Particle System

If there are more than two particles in a system, then the net gravitational potential energy of the whole system is the sum of gravitational potential energies of all the possible pairs in that system.

$U_{System} = (G \frac{m _{1} m _{2}}{r _{1}}) + (G \frac{m _{2} m _{3}}{r _{2}}) + (G \frac{m _{1} m _{3}}{r _{3}})$

$= - G \frac{m _{1} m _{2}}{r _{1}} - G \frac{m _{2} m _{3}}{r _{2}} - G \frac{m _{1} m _{3}}{r _{3}}$

If Gravitational Potential at a point P is V_P
Then GPE of a particle of mass m placed at P is :U_P = m(V_P)

Note: At infinity, $V_{\infty} = 0 \Rightarrow U_{\infty} = 0$

6.0Escape Velocity

It is the minimum velocity required for an object located at the planet's surface so that it just escapes the planet's gravitational field

$V_{e} = \frac{2 GM}{R _{e}}$

$V_{e} = 2 g_{s} R_{e} = 11.2 km / s . R_{e} = 6400 km$

Escape speed depends on :

Mass (M), radius (R) of the planet
Position from where the particle is projected.

Escape speed does not depend on :

Mass (m) of the body which is projected
Angle of projection.
If a body is thrown from the Earth's surface with escape speed, it goes out of earth's gravitational field and never returns back to the earth's surface.

Escape velocity from a point other than surface

1. Total energy is zero at any point when particle start moving with escape velocity,T E=0

2. If given point is at distance r (>R) from center of Earth, $P E_{Out} = - \frac{G M _{e} m}{r}$

3. If given point is at distance r (<R) from the center of Earth, $P E_{I n} = - \frac{GM ( 3 R ^{2} - r ^{2} )}{2 R ^{3}} \times m$

7.0Escape Energy-Binding Energy

Minimum energy given to a particle in the form of kinetic energy so that it can just escape the Earth's gravitational field.
Magnitude of escape energy = $G \frac{M m}{R}$
(–ve of PE on the Earth's surface)
Escape energy = Kinetic Energy corresponding to the escape velocity

$G \frac{M m}{R} = \frac{1}{2} m v_{e}^{2}$

KE < Escape Energy v_o < v_e Body returns to Earth surface
KE = Escape Energy v_o = v_e Body comes to rest at infinity
KE > Escape Energy v_o > v_e Body has residual velocity at infinity

Binding energy:

Total energy of a particle near Earth.

BE < 0 Particles cannot escape the gravitational field of Earth.
BE ≥ 0 Particles can escape the gravitational field of Earth.

8.0Kepler’s Law

Kepler's $1^{s t}$ Law: Each planet moves in an elliptical revolution, with the sun at one focus of the ellipse.

Kepler's $2^{n d}$ Law: A line joining any planet to the Sun glides out equal areas in equal intervals of time, i.e. the areal velocity of the planet remains constant.

Kepler's $3^{r d}$ Law: The square of the period of revolution of any planet around the Sun is directly relationall to the cube of the semi-major axis of its elliptical orbit.

$T^{2} \propto a^{3}$

a : average distance of the planet from Sun

$a = \frac{r _{m a x} + r _{m i n}}{2}$

9.0Satellite Motion

A light body revolving around a heavier planet due to gravitational attraction, is called a satellite. Moon is a natural satellite of Earth.

Essential conditions for satellite motion:

The centre of the satellite's orbit should coincide with the centre of Earth
Plane of the orbit of the satellite should pass through the centre of Earth.

It follows that a satellite can revolve around the earth only in those circular orbits whose centres coincide with the centre of earth. Circles drawn on a globe with centres coincident with earth are known as 'great circles'. Therefore, a satellite revolves around the earth along circles concentric with great circles.

Also Read : Earth Satellites

10.0Orbital Velocity (V_o)

$V_{o} = \frac{G M _{e}}{R _{e}} \approx 8.1 km / s$

Time period of orbital motion:

$T = 2 π \frac{R _{e}^{3}}{G M _{e}} \approx 84 min$

Satellite motion - Energy Analysis

1.Kinetic energy of satellite

$K E = \frac{G M _{e} m}{2 r}$

2.Potential energy of satellite

$PE = - \frac{G M _{e} m}{r}$

Total Energy = Kinetic Energy + Potential Energy

$TE = - \frac{G M _{e} m}{2 r}$

$K E = - TE = - \frac{PE}{2} = - \frac{G M _{e} m}{2 r}$

As total energy of a satellite is negative, it is called as bounded system

Also Read : Kepler's Law of Planetary Motion

11.0Binding Energy

Minimum energy required to break a bounded system.
$BE + TE = 0 \Rightarrow BE = - TE$
$BE = ∣ TE ∣ \Rightarrow BE = \frac{G M _{e} m}{2 r}$

Note:

$TE = - \frac{G M _{e} m}{2 r}$
$PE = - \frac{G M _{e} m}{r}$
$K E = \frac{G M _{e} m}{2 r}$
$V_{0} = \frac{G M _{e}}{r}$

$\Rightarrow T \propto R^{3/2}$

So, If the radius of the orbit of the satellite is increased,

KE will decrease, TE will increase, PE will increase, Orbital velocity will decrease, and Time period will increase.

12.0Work Done to Shift Orbit of Satellite

$W_{E x t} = Δ TE$

$w_{E x t} = \frac{G M _{ε} m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$

Essential conditions for satellite motion

The Center of the satellite's orbit coincides with the center of Earth.
Plane of orbit of the satellite is passing through the center of Earth.

There are two stable orbits for a satellite

Equatorial orbit
Polar orbit

Any satellite is projected as near to the equator as possible as-

g is minimum near the equator, less fuel
Easier to set the satellite in any orbit

13.0Polar Satellite

It rotates in the Polar plane.

Its height from Earth's surface is 500 km ~ 600 km.
Its time period is 100 min.
Used for weather data collection

Note: Any satellite is said to be in a condition of freely falling on Earth. (as the only force acting on it is the force of gravity). Hence, the apparent weight of any object in a satellite is always ZERO.

Related Video:

Also Read:

Escape Velocity	Gravitational Potential Energy	Potential Energy
Kepler's Law of Planetary Motion	Acceleration Due To Gravity	Kinetic Energy
Vectors in Physics	Centre of Mass And Collision	Elastic Collision

Frequently Asked Questions

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(Session 2026 - 27)

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Gravitation: Download Important IIT JEE & NEET Questions

Gravitation is a fundamental force of nature that governs the motion of objects throughout the universe. It is the attractive force that draws two bodies toward one another, resulting in a pull that depends on their masses and the distance separating them. This force not only keeps us anchored to the Earth but also plays a vital role in the movement of celestial bodies, such as planets, moons, and stars.

The understanding of gravitation has significantly evolved over the centuries. In the late 17th century, Sir Isaac Newton was among the first to mathematically describe this force through his Universal Law of Gravitation.

1.0Newton Law of Gravitation

Gravitation is the property of matter due to which two bodies always attract each other.
It states that every particle in the universe attracts all other particles with a force which is directly commensurate to the product of their masses and is inversely commensurate to the square of the distance between them.
This formula is given for point masses. For spherical objects (planets), the distance between centres is taken.

$F_{g} = G \frac{m _{1} m _{2}}{r ^{2}}$

G=universal Gravitational Constant,Its value is the same throughout the universe; G independent of the nature and size of the bodies; it does not depend even upon the nature of the medium between the bodies.
Value of G in $SI = 6.67 \times 1 0^{- 11} Nm^{2} kg^{- 2} or CG.S = 6.67 \times 1 0^{- 8} dyne cm^{2} g^{- 2}$
Dimension of $G = [M^{- 1} L^{3} T^{- 2}]$
Forces due to Multiple Particles is given as, $F_{n e t} = F_{1} + F_{2} + F_{3} + \dots\dots$

Law of Gravitation in Vector Form

$F_{12} = - G \frac{m _{1} m _{2}}{r ^{2}} \overset{r}{^}$

Related Video:

2.0Gravitational Field

It is the region around a mass where its gravitational effects can be observed. The gravitational field due to a body extends till infinity.
Strength of the gravitational field decreases on moving away from a body.

(i) In an infinite gravitational field, the acceleration is zero.

(ii) Gravitational field ∝ mass

Gravitational field is conservative in nature.
There are two ways to measure the strength of the gravitational field.

(i) Gravitational field intensity.

(ii) Gravitational potential

Gravitational Field Intensity

It is defined as gravitational force exerted on a unit mass positioned in a gravitational field.
It is a vector quantity.Its direction is same as that of gravitational force.
$I = \frac{F}{m} \Rightarrow Unit - N / K g$
Dimensional Formula $- [M^{0} L^{1} T^{- 2}]$
Gravitational Field Intensity owed to a point mass,is given as,

$I = \frac{GM}{r ^{2}} (- \overset{r}{^})$

Negative sign indicates that the force due to the gravitational field is attractive in nature.
Gravitational Field Intensity due to multiple particles,

$I_{n e t} = I_{1} + I_{2} + I_{3} + \dots\dots\dots\dots\dots$

A. Solid Sphere

Outside the Sphere (r>R)

$I_{out} = \frac{GM}{r ^{2}} (- \overset{r}{^})$

At the Surface (r=R)

$I_{surface} = \frac{GM}{R ^{2}} (- r)$

Inside the Sphere (r<R)

$I_{in} = \frac{GM r}{r ^{3}} (- \overset{r}{^})$

Graph between $I$ and r for a solid sphere

B. Hollow Sphere

Outside the Shell (r>R)

$I_{out} = \frac{GM}{r ^{2}} (- \overset{r}{^})$

At the Surface (r=R)

$I_{surface} = \frac{GM}{R ^{2}} (- \overset{r}{^})$

Inside the Shell (r<R) I=0

Graph between $I$ and r for Hollow sphere

3.0Acceleration Due to Gravity

If a body is placed inside a gravitational field having intensity I, gravitational force applied on it is, F=m I
Earth also pulls every object towards its center with force F=mg

On comparing both the equations we get $g = ∣ I ∣$

Where g = acceleration due to gravity I is intensity of gravitational field due to Earth.

Assuming Earth to be a Solid Sphere

$g_{s} = \frac{GM}{R _{e}^{2}} \Rightarrow 9.81 m / s^{2}$ , Same formula can be used to find acceleration due to gravity at the surface of any planet.

In terms of density $(ρ) : g = \frac{GM}{R ^{2}}$

$mass = density \times Volume$

$M = ρ \times \frac{4}{3} π R^{3}$

$g = \frac{G ( \frac{4}{3} π R ^{3} )}{R ^{2}} = \frac{4}{3} π GRρ$

Variation in acceleration due to gravity with height

$g_{h} = g [1 - \frac{2 h}{R _{e}}]$ , This formula is valid if h is upto 5% of earth's radius (320 km from earth's surface)

If h is greater than 5% earth's radius we use, $g_{h} = \frac{G M _{e}}{( R _{e} + h ) ^{2}}$

Variation in acceleration due to gravity with depth

$g_{d} = g_{s} (1 - \frac{d}{R})$ , For any depth below Earth's Surface

Note:

Valid for small or large depth.
Fractional change in value of $g \to \frac{Δ g}{g} = \frac{d}{R}$
At any depth, $g_{d} < g_{s}$
Another form that can be used $g_{d} = \frac{GM}{R ^{3}} r \to g_{d} = g_{s} \frac{r}{R}$

Graph

For (r<R)

$g = \frac{GM}{R ^{3}} r \to g = g_{s} \frac{r}{R}$

For $(r \geq R)$

$g = \frac{GM}{r ^{2}} \to g = g_{s} \frac{R ^{2}}{r ^{2}}$

Value of g is maximum at Earth's surface

Variation in value of 'g' due to rotation of earth

$N = M g_{e ff} = M g - M ω^{2} R_{e} Cos^{2} λ$ , As one moves towards pole, apparent weight increases

$λ : Angle of Latitude$

$For Poles λ = 9 0^{\circ}$

$For Equator λ = 0^{\circ}$

$g =$ acceleration due to gravity without considering rotation of Earth

$g_{eff} =$ acceleration due to gravity considering rotation of Earth

Variation in value of "g" due to shape of Earth

$Δ g = g_{e} - g_{p} \approx 0.02 m / s^{2}$ , Earth is not a perfect sphere

$g_{surface} = \frac{GM}{R ^{2}}$

$g_{p} > g_{eq’}$ (as surface is close to center of Earth near pole as compared to equator )

4.0Gravitational Potential

It is defined as amount of work done by external agent against gravitational pull in conveying a unit mass from infinity to that point without changing its kinetic energy

$V_{P} = \frac{W _{\infty \to p}}{m}$

Motion of unit mass is considered slowly, so that no energy is spent to change the kinetic energy of the system.
At infinity, Gravitational Potential is assumed to be ZERO.

Also Read : Difference Between Mass and Weight

Gravitational Potential Due To a Point Mass:

$V = \frac{W _{e x t}}{m} = - \frac{GM}{r}$ (Negative sign in potential indicates attractive nature of force. )

Relation between Potential and Intensity

$I = - \frac{d V}{d r}$

If V is constant in a region, I = 0
This relation is valid for all conservative fields.

Standard results to remember:

Spherical shell (M, R)

Case 1. Outside the Shell $(r > R), V_{out} = - \frac{GM}{r}$

Case 2. On the Surface $(r = R), V_{Surface} = - \frac{GM}{R}$

Case 3. Inside the Shell $(r < R), V_{I n} = - \frac{GM}{R}$ , (Potential is same everywhere and is equal to its value at the surface.)

Solid sphere(M, R)

Case 1. Outside the Sphere $(r > R), V_{out} = - \frac{GM}{r}$

Case 2. On the Surface $(r = R), V_{Surface} = - \frac{GM}{R}$

Case 3. Inside the Sphere $(r < R), V_{I n} = - \frac{GM}{2 R ^{3}} (3 R^{2} - r^{2})$ , (Potential is same everywhere and is equal to its value at the surface.)

5.0Gravitational Potential Energy

The gravitational potential energy of a particle situated at a point in some gravitational field is defined as the amount of work required to bring it from infinity to that point without changing its kinetic energy.

$U = - G \frac{m _{1} m _{2}}{r}$ , negative sign shows the boundedness of the two bodies

It is a scalar quantity.
It's SI unit is joule and Dimensions are [M¹L²T^–2]
The gravitational potential energy of a particle of mass 'm' placed on the surface of earth of mass 'M' , radius 'R' is given by : $U = - G \frac{m _{1} m _{2}}{R}$

Gravitational Potential Energy for Three Particle System

If there are more than two particles in a system, then the net gravitational potential energy of the whole system is the sum of gravitational potential energies of all the possible pairs in that system.

$U_{System} = (G \frac{m _{1} m _{2}}{r _{1}}) + (G \frac{m _{2} m _{3}}{r _{2}}) + (G \frac{m _{1} m _{3}}{r _{3}})$

$= - G \frac{m _{1} m _{2}}{r _{1}} - G \frac{m _{2} m _{3}}{r _{2}} - G \frac{m _{1} m _{3}}{r _{3}}$

If Gravitational Potential at a point P is V_P
Then GPE of a particle of mass m placed at P is :U_P = m(V_P)

Note: At infinity, $V_{\infty} = 0 \Rightarrow U_{\infty} = 0$

6.0Escape Velocity

It is the minimum velocity required for an object located at the planet's surface so that it just escapes the planet's gravitational field

$V_{e} = \frac{2 GM}{R _{e}}$

$V_{e} = 2 g_{s} R_{e} = 11.2 km / s . R_{e} = 6400 km$

Escape speed depends on :

Mass (M), radius (R) of the planet
Position from where the particle is projected.

Escape speed does not depend on :

Mass (m) of the body which is projected
Angle of projection.
If a body is thrown from the Earth's surface with escape speed, it goes out of earth's gravitational field and never returns back to the earth's surface.

Escape velocity from a point other than surface

1. Total energy is zero at any point when particle start moving with escape velocity,T E=0

2. If given point is at distance r (>R) from center of Earth, $P E_{Out} = - \frac{G M _{e} m}{r}$

3. If given point is at distance r (<R) from the center of Earth, $P E_{I n} = - \frac{GM ( 3 R ^{2} - r ^{2} )}{2 R ^{3}} \times m$

7.0Escape Energy-Binding Energy

Minimum energy given to a particle in the form of kinetic energy so that it can just escape the Earth's gravitational field.
Magnitude of escape energy = $G \frac{M m}{R}$
(–ve of PE on the Earth's surface)
Escape energy = Kinetic Energy corresponding to the escape velocity

$G \frac{M m}{R} = \frac{1}{2} m v_{e}^{2}$

KE < Escape Energy v_o < v_e Body returns to Earth surface
KE = Escape Energy v_o = v_e Body comes to rest at infinity
KE > Escape Energy v_o > v_e Body has residual velocity at infinity

Binding energy:

Total energy of a particle near Earth.

BE < 0 Particles cannot escape the gravitational field of Earth.
BE ≥ 0 Particles can escape the gravitational field of Earth.

8.0Kepler’s Law

Kepler's $1^{s t}$ Law: Each planet moves in an elliptical revolution, with the sun at one focus of the ellipse.

Kepler's $2^{n d}$ Law: A line joining any planet to the Sun glides out equal areas in equal intervals of time, i.e. the areal velocity of the planet remains constant.

Kepler's $3^{r d}$ Law: The square of the period of revolution of any planet around the Sun is directly relationall to the cube of the semi-major axis of its elliptical orbit.

$T^{2} \propto a^{3}$

a : average distance of the planet from Sun

$a = \frac{r _{m a x} + r _{m i n}}{2}$

9.0Satellite Motion

A light body revolving around a heavier planet due to gravitational attraction, is called a satellite. Moon is a natural satellite of Earth.

Essential conditions for satellite motion:

The centre of the satellite's orbit should coincide with the centre of Earth
Plane of the orbit of the satellite should pass through the centre of Earth.

It follows that a satellite can revolve around the earth only in those circular orbits whose centres coincide with the centre of earth. Circles drawn on a globe with centres coincident with earth are known as 'great circles'. Therefore, a satellite revolves around the earth along circles concentric with great circles.

Also Read : Earth Satellites

10.0Orbital Velocity (V_o)

$V_{o} = \frac{G M _{e}}{R _{e}} \approx 8.1 km / s$

Time period of orbital motion:

$T = 2 π \frac{R _{e}^{3}}{G M _{e}} \approx 84 min$

Satellite motion - Energy Analysis

1.Kinetic energy of satellite

$K E = \frac{G M _{e} m}{2 r}$

2.Potential energy of satellite

$PE = - \frac{G M _{e} m}{r}$

Total Energy = Kinetic Energy + Potential Energy

$TE = - \frac{G M _{e} m}{2 r}$

$K E = - TE = - \frac{PE}{2} = - \frac{G M _{e} m}{2 r}$

As total energy of a satellite is negative, it is called as bounded system

Also Read : Kepler's Law of Planetary Motion

11.0Binding Energy

Minimum energy required to break a bounded system.
$BE + TE = 0 \Rightarrow BE = - TE$
$BE = ∣ TE ∣ \Rightarrow BE = \frac{G M _{e} m}{2 r}$

Note:

$TE = - \frac{G M _{e} m}{2 r}$
$PE = - \frac{G M _{e} m}{r}$
$K E = \frac{G M _{e} m}{2 r}$
$V_{0} = \frac{G M _{e}}{r}$

$\Rightarrow T \propto R^{3/2}$

So, If the radius of the orbit of the satellite is increased,

KE will decrease, TE will increase, PE will increase, Orbital velocity will decrease, and Time period will increase.

12.0Work Done to Shift Orbit of Satellite

$W_{E x t} = Δ TE$

$w_{E x t} = \frac{G M _{ε} m}{2} (\frac{1}{r _{1}} - \frac{1}{r _{2}})$

Essential conditions for satellite motion

The Center of the satellite's orbit coincides with the center of Earth.
Plane of orbit of the satellite is passing through the center of Earth.

There are two stable orbits for a satellite

Equatorial orbit
Polar orbit

Any satellite is projected as near to the equator as possible as-

g is minimum near the equator, less fuel
Easier to set the satellite in any orbit

13.0Polar Satellite

It rotates in the Polar plane.

Its height from Earth's surface is 500 km ~ 600 km.
Its time period is 100 min.
Used for weather data collection

Note: Any satellite is said to be in a condition of freely falling on Earth. (as the only force acting on it is the force of gravity). Hence, the apparent weight of any object in a satellite is always ZERO.

Related Video:

Also Read:

Escape Velocity	Gravitational Potential Energy	Potential Energy
Kepler's Law of Planetary Motion	Acceleration Due To Gravity	Kinetic Energy
Vectors in Physics	Centre of Mass And Collision	Elastic Collision

Frequently Asked Questions

Why is gravitational potential energy typically represented as a negative value?

What are two key factors that influence whether a planet can sustain an atmosphere?

Why does a body experience weightlessness at the center of the Earth?

Write characteristics of Gravitational force?

What fundamental law of physics does Kepler's second law represent?

Join ALLEN!

Gravitation: Download Important IIT JEE & NEET Questions

1.0Newton Law of Gravitation

Law of Gravitation in Vector Form

2.0Gravitational Field

Gravitational Field Intensity

3.0Acceleration Due to Gravity

4.0Gravitational Potential

5.0Gravitational Potential Energy

Gravitational Potential Energy for Three Particle System

6.0Escape Velocity

7.0Escape Energy-Binding Energy

8.0Kepler’s Law

9.0Satellite Motion

10.0Orbital Velocity (Vo)

Satellite motion - Energy Analysis

11.0Binding Energy

12.0Work Done to Shift Orbit of Satellite

13.0Polar Satellite

Table of Contents

Frequently Asked Questions

Why is gravitational potential energy typically represented as a negative value?

What are two key factors that influence whether a planet can sustain an atmosphere?

Why does a body experience weightlessness at the center of the Earth?

Write characteristics of Gravitational force?

What fundamental law of physics does Kepler's second law represent?

Join ALLEN!

Gravitation: Download Important IIT JEE & NEET Questions

1.0Newton Law of Gravitation

Law of Gravitation in Vector Form

2.0Gravitational Field

Gravitational Field Intensity

3.0Acceleration Due to Gravity

4.0Gravitational Potential

5.0Gravitational Potential Energy

Gravitational Potential Energy for Three Particle System

6.0Escape Velocity

7.0Escape Energy-Binding Energy

8.0Kepler’s Law

9.0Satellite Motion

10.0Orbital Velocity (Vo)

Satellite motion - Energy Analysis

11.0Binding Energy

12.0Work Done to Shift Orbit of Satellite

13.0Polar Satellite

Table of Contents

10.0Orbital Velocity (V_o)

10.0Orbital Velocity (V_o)