Gravitational Potential & Relation With Gravitational Potential Energy-Derivation, Formula, Examples | JEE NEET CBSE Class 11 Notes

Introduction to Gravitational Potential

Gravitational potential is a fundamental concept in physics that describes the potential energy per unit mass at a point in a gravitational field. It plays a crucial role in solving problems related to gravitational force, energy, and planetary motion. Understanding gravitational potential is essential for students preparing for JEE, NEET, and CBSE Class 11 exams.

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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.

Gravitational potential at a point is defined as the work done per unit mass in bringing a test mass from infinity to that point in a gravitational field.

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

⇒ V = U/m

Mathematically, it is given by:

V = -GM/r

Where :

• V = Gravitational potential
• G = Universal Gravitational Constant (6.674 × 10⁻¹¹ Nm²/kg²)
• M = Mass of the body creating the gravitational field
• r = Distance from the center of the mass

Gravitational Potential increases (becomes less negative) as the distance increases because the gravitational field weakens with distance.

⇒ 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/Kg.
• The dimensional formula is [M⁰L²T^-2].

Characteristics of Gravitational Potential (V)

• It is a scalar quantity.
• The value is always negative because work is done against the gravitational force.
• Gravitational potential is zero (maximum) at infinity.
• The potential at a point inside a uniform sphere is constant.

Relation between Gravitational Field Intensity and Gravitational Potential

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

\(\begin{array}{l}V = -\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

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

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

E = -dV/dr

\(\begin{array}{l}\overrightarrow{~E}=\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\end{array} \)

(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 increases (becomes less negative) as the distance increases because the gravitational field weakens with distance

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 inside the spherical shell.

E = -dV/dr

-dV/dr = 0

V = Constant

So, Gravitational Potential is constant inside a Spherical Shell.

The value of gravitational potential inside a Spherical Shell 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/R².

Using the relation

\(\begin{array}{l}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

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

Gravitational Potential (V) = -GM/R.

The value of gravitational potential at the surface of Spherical Shell is given by,

V = -GM/R

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

Outside the spherical shell, E = -GM/r².

Using the relation

\(\begin{array}{l}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

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

V = -GM/r

Gravitational Potential increases (becomes less negative) as the distance increases because the gravitational field weakens with distance

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/R³.

Using the relation

\(\begin{array}{l}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

over a limit of (0 to r).

The value of gravitational potential is given by,

V = -GM [(3R² – r²)/2R²]

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/R².

Using the relation

\(\begin{array}{l}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} \)

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.,

\(\begin{array}{l}{{U}_{s}}=\frac{1}{2}Gn\left( n-1 \right)\frac{{{m}^{2}}}{{{r}^{2}}}\end{array} \)

Solved Problems

Question 1: What is the gravitational potential of the sun with respect to the earth. The mass of the sun is 1.99×10³⁰ kgs. The distance between the sun and earth is 150×10⁶ km?

Answer:

Given that,

Mass of sun M = 1.99×10³⁰kg.

Distance r = 150×10⁶ km.

Gravitational Potential V = – GM/r

V = (-6.67×10^-11×1.99×10³⁰)/ (150×10⁶)

= 8.85×10⁸ J/kg.

Therefore, the gravitational potential of sun with respect to earth is 8.85×10⁸ J/kg.

Question 2: What is the Gravitational Potential of an object with of mass 2×10²⁴ kg with respect to another object. It is at a distance of 1400 km.

Answer:

Given that,

Mass of object M = 2×10²⁴ kg.

Distance r = 1400 km.

Gravitational Potential V = – GM/r

(-6.67×10^-11×2× 10²⁴) / (1400×1000)

V = 9.52×10⁷ J/kg.

Therefore, the gravitational potential required is 9.52×10⁷ J/kg.

Question 3: Where is gravitational potential minimum. What is it minimum value?

Answer:

Gravitational potential is minimum at the centre of the earth. The minimum value of gravitational potential is not defined.

Question 4: What is the Gravitational Potential Energy of a body at a distance of 3r from the center of the earth.

Answer:

Given that,

Distance r = 3r

We know that, gravitational potential energy at distance r is 

U = -GMm/r

Gravitational Potential Energy at distance 3r is

= – GMm/3r

= U/3

Therefore, gravitational potential energy at distance 3r is U/3.

Question 5: The Gravitational Potential Energy of a body at a distance of 2r from the earth is U. Find the gravitational force at that point.

Answer:

Given that,

Gravitational Potential Energy = U

distance r = 2r

We know that, gravitational potential energy at distance r is 

U = GMm/r

Formula of gravitational force F = GMm/r²

F = GMm/(2r)²

= GMm/4r²

= GMm/(4r)r

= U/4r

Therefore, the gravitational force at distance 2r is U/4r.

Multiple-Choice Questions (MCQs) on Gravitational Potential

Q1: Which of the following is correct about gravitational potential?

A) It is a vector quantity
B) It is always positive
C) It is always negative
D) It is infinite at a point

Correct Answer: C) It is always negative
Explanation: Gravitational potential is negative because work is done against the gravitational field to bring a mass from infinity to a given point.

Q2: The gravitational potential V at a distance r from a mass M is given by V = -GM/r . What happens to the potential if the distance is doubled?

A) It remains the same
B) It doubles
C) It halves
D) It becomes one-fourth

Correct Answer: C) It halves
Explanation: Since gravitational potential varies inversely with distance, doubling r results in half the potential value.

Frequently Asked Questions (FAQs) on Gravitational Potential Energy

Q1: Why is gravitational potential negative?

A: It is negative because work is required to bring an object from infinity (where potential is taken as zero) to a point in a gravitational field.

Q2: What is the difference between gravitational potential and gravitational potential energy?

A: Gravitational potential is the potential energy per unit mass, while gravitational potential energy is the total energy stored due to an object’s position in a gravitational field.

Q3: How does gravitational potential affect satellite motion?

A: Satellites orbit due to the balance between gravitational attraction and their tangential velocity. The potential determines the total energy required to maintain an orbit.

Q4: How does gravitational potential change with distance?

A: It increases (becomes less negative) as the distance increases because the gravitational field weakens with distance.

Q5: What is gravitational potential?

A: The gravitational potential at a point in the gravitational field of a body is defined as the amount of work done in displacing a body of unit mass from infinity to that point in the field. It is denoted as V.

Q6: What is the expression for gravitational potential energy?

A: Gravitational potential Energy, U = -GMm/r
M is the source mass placed along the x-axis.
m is the test mass at infinity.
r is the distance from the source mass at which the gravitational potential energy is determined.

Q7: What is the unit of gravitational potential?

A: The unit of gravitational potential is J/kg.

Q8: What is the dimensional formula of gravitational potential?

A: The dimensional formula of gravitational potential is [M⁰L²T^-2]

Applications of Gravitational Potential

• Satellite motion and orbital mechanics
• Calculation of escape velocity
• Energy calculations in astrophysics
• Black hole and planetary studies

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