Gravitation for NEET: Universal Law, Kepler’s Laws, Escape Velocity and Satellite Motion

Gravitation is the force of attraction between any two masses in the universe. Every object attracts every other object, although in most everyday cases the force is too small to notice. Gravitation explains why objects fall towards Earth, why planets move around the Sun, and why satellites remain in orbit.

What is Gravitational Force?

Gravitational force is the force with which two masses attract each other. It is always attractive, always acts along the line joining the centres of the two bodies, and depends only on the masses and the distance between them. 

Universal Law of Gravitation

According to the universal law of gravitation, every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.

Mathematically,

F = Gm1m2 / r²

where
F = gravitational force
G = universal gravitational constant
m1 and m2 = masses of the two bodies
r = distance between their centres

Value and SI Unit of Gravitational Constant

The universal gravitational constant is

G = 6.67 × 10⁻¹¹ N m² kg⁻²

Its dimensional formula is

[M⁻¹ L³ T⁻²]

Vector Form of Newton’s Law of Gravitation

The gravitational force between two bodies is directed along the line joining them. If r-hat is the unit vector from one mass to the other, then the force can be written in vector form with a negative sign to indicate attraction.

This vector treatment is useful in understanding direction and in proving that gravitational forces form an action-reaction pair.    

Properties of Gravitational Force

• Gravitational force is always attractive.

• It is a central force.

• It is a conservative force.

• It obeys the inverse square law.

• It has an infinite range.

• It follows the principle of superposition.

These properties are directly useful in NEET theory questions.

Principle of Superposition of Gravitation

If many masses act on a body, the net gravitational force is the vector sum of all individual gravitational forces acting on that body.

Fnet = F1 + F2 + F3 + …

Acceleration Due to Gravity

Acceleration due to gravity is the acceleration produced in a body due to Earth’s gravitational pull. It is denoted by g.

At the surface of Earth,   

g = GM / R²

where
M = mass of Earth
R = radius of Earth

Near Earth’s surface, the standard value is approximately

g = 9.8 m/s²

This relation follows directly from Newton’s law of gravitation and the idea that the gravitational force on a body equals its weight.

Relation Between g and G

For a body of mass m on Earth’s surface,

mg = GMm / R²

So,

g = GM / R²

This relation is very important for deriving formulas involving height and depth.

Variation of g with Altitude

At a height h above Earth’s surface,

g(h) = GM / (R + h)²

This means acceleration due to gravity decreases as altitude increases.

For h much smaller than R,

g(h) ≈ g(1 - 2h/R)

Key point: As we go higher above Earth, g decreases.

Variation of g with Depth

At a depth d below Earth’s surface,

g(d) = g(1 - d/R)

This means acceleration due to gravity decreases linearly with depth, assuming Earth has uniform density.

At the centre of Earth,

g = 0

Key point: g is maximum at Earth’s surface and decreases both with height and depth.

Graph of g with Altitude and Depth

With altitude, g decreases non-linearly.

With depth, g decreases linearly and becomes zero at the centre.

These trend-based ideas are often asked in NEET conceptual questions.

Gravitational Potential Energy

The gravitational potential energy of a body in the gravitational field of another body is the energy due to its position.

For two masses separated by distance r,

U = -GMm / r

The negative sign shows that the gravitational force is attractive, and work must be done to separate the masses to infinity.

Gravitational Potential

Gravitational potential at a point is defined as the potential energy per unit mass at that point.

V = U / m = -GM / r

It is a scalar quantity and is always negative for an attractive gravitational field.

Difference Between Potential and Potential Energy

• Gravitational potential is potential energy per unit mass.

• Gravitational potential energy depends on both interacting masses.

• Potential is a property of the field at a point.

• Potential energy is the energy of a particular body placed in that field. 

Kepler’s Laws of Planetary Motion

Kepler’s First Law

Each planet moves around the Sun in an elliptical orbit with the Sun at one focus.

Kepler’s Second Law

The line joining the Sun and a planet sweeps out equal areas in equal intervals of time.

Kepler’s Third Law

The square of the time period of revolution of a planet is directly proportional to the cube of the semi-major axis of its orbit.

T² ∝ r³

Importance of Kepler’s Laws for NEET   

Kepler’s laws explain planetary motion clearly and help derive relations involving orbital time period and distance from the Sun. They are especially important for understanding satellite motion and gravitation-based orbital questions. 

Escape Velocity   

Escape velocity is the minimum speed required for a body to escape completely from the gravitational field of a planet without further propulsion.

For Earth,

ve = √(2GM / R)

Using g = GM/R²,

ve = √(2gR)

For Earth, its approximate value is 11.2 km/s.

Key point: Escape velocity does not depend on the mass of the escaping body.

Orbital Velocity

Orbital velocity is the minimum horizontal speed required by a satellite to remain in a stable circular orbit around a planet.

For a satellite near Earth,

vo = √(GM / r)

For an orbit very close to Earth’s surface,

vo = √(GM / R) = √(gR)

For Earth, this is about 7.9 km/s.

Relation Between Escape Velocity and Orbital Velocity

ve = √2 vo

This is a very important direct result relation for NEET.

Motion of a Satellite

A satellite remains in orbit because Earth’s gravitational force provides the required centripetal force.

GMm / r² = mv² / r

From this, we get orbital velocity:

v = √(GM / r)

Time Period of a Satellite

For a satellite in a circular orbit of radius r,

T = 2π √(r³ / GM)

This directly shows that the time period increases with orbital radius.

Energy of a Satellite

The total mechanical energy of a satellite in circular orbit is

E = K + U

Kinetic energy,

K = GMm / 2r

Potential energy,

U = -GMm / r

Total energy,

E = -GMm / 2r

Important result:

Total energy is negative, which means the satellite is bound to the planet.

Important Gravitation Formulas for NEET

Gravitational force:
F = Gm1m2 / r²

Acceleration due to gravity:
g = GM / R²

g at height h:
g(h) = GM / (R + h)²

Approximation:
g(h) ≈ g(1 - 2h/R)

g at depth d:
g(d) = g(1 - d/R)

Potential energy:
U = -GMm / r

Potential:
V = -GM / r

Escape velocity:
ve = √(2GM / R) = √(2gR)

Orbital velocity:
vo = √(GM / r)

Satellite time period:
T = 2π √(r³ / GM)

Satellite total energy:
E = -GMm / 2r

Kepler’s third law:
T² ∝ r³

Comparison of Escape Velocity and Orbital Velocity

• Orbital velocity is the speed needed to keep a body in orbit.

• Escape velocity is the speed required to escape a gravitational field.

• Escape velocity is greater than orbital velocity.

• For Earth, orbital velocity is about 7.9 km/s, and escape velocity is about 11.2 km/s.

Common Mistakes Students Make in Gravitation

• Students often confuse g with G.

• They forget that G is universal, but g changes from place to place.

• They use R instead of R + h in altitude questions.

• They forget the negative sign in gravitational potential and potential energy.

• They mix up orbital velocity and escape velocity.

• They use satellite formulas for bodies that are not in circular orbits.

• They forget that gravitational potential is a scalar, while force is a vector.

Solved Examples on Gravitation

Example 1

Find the gravitational force between two bodies of masses 40 kg and 60 kg kept 2 m apart in air.

Solution:

Given,
m1 = 40 kg
m2 = 60 kg
r = 2 m
G = 6.67 × 10⁻¹¹ N m² kg⁻²

Using,
F = Gm1m2 / r²

F = (6.67 × 10⁻¹¹ × 40 × 60) / 4

F = 4.002 × 10⁻⁸ N

Answer:
The gravitational force is 4.002 × 10⁻⁸ N.

Example 2

Calculate the value of g at a height equal to Earth’s radius above the surface. Let g at the surface be 9.8 m/s².

Solution:

Here,
h = R

Using,
g(h) = g × R² / (R + h)²

g(h) = 9.8 × R² / (2R)²
= 9.8 / 4
= 2.45 m/s²

Answer:
The acceleration due to gravity at that height is 2.45 m/s².

Example 3

Find the gravitational potential at the surface of a planet of mass 8 × 10²⁴ kg and radius 8 × 10⁶ m. 

Solution:

Using,
V = -GM / R

V = -(6.67 × 10⁻¹¹ × 8 × 10²⁴) / (8 × 10⁶)

V = -6.67 × 10⁷ J/kg

Answer:
The gravitational potential is -6.67 × 10⁷ J/kg.

Example 4

A satellite revolves close to Earth’s surface. Find its orbital speed if g = 9.8 m/s² and Earth’s radius is 6.4 × 10⁶ m.

Solution:

Using,
v = √(gR)

v = √(9.8 × 6.4 × 10⁶)
= √(6.272 × 10⁷)
≈ 7.92 × 10³ m/s

Answer:
The orbital speed is about 7.92 km/s.

Example 5

Find the escape velocity from a planet where g = 12 m/s² and radius = 5 × 10⁶ m.

Solution:

Using,
ve = √(2gR)

ve = √(2 × 12 × 5 × 10⁶)
= √(1.2 × 10⁸)
≈ 1.095 × 10⁴ m/s

Answer:
The escape velocity is about 10.95 km/s.

Example 6

A satellite moves in a circular orbit of radius 4R, where R is Earth’s radius. If the time period of a satellite near Earth’s surface is T, what will be the new time period?

Solution:

Using the Kepler-type relation for a circular orbit,
T² ∝ r³

So,

Tnew / T = (4R / R)^(3/2)
= 4^(3/2)
= 8

Therefore,
Tnew = 8T

Answer:
The new time period is 8T.

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