Q.38 Consider two bodies with equal masses of 1012 kg each and R distance apart. Let G be the
gravitational constant and V0 be a constant with dimensions of energy. Which of the
following represent(s) gravitational potential energy (V) between the bodies, such that
Newton’s law of gravitation is valid?
The correct expressions for the gravitational potential energy between the two bodies are options (A) and (B).
Problem Restatement
Two bodies each have mass 1012 kg and are separated by distance R. Newton’s law of gravitation gives the magnitude of the force as
F = G m1 m2 / R2, or F = R-2 G m1 m2,
where G is the gravitational constant and m1, m2 are the interacting masses.
The question asks which given expressions can represent the gravitational potential energy V(R) between the bodies such that the force obtained from V matches Newton’s law.
For a central potential depending only on R, the radial force is
F(R) = - dV/dR.
Here m1 = m2 = 1012 kg, so m1 m2 = 1024 kg2.
Deriving the Correct Gravitational Potential Energy
For gravity between two point masses, the standard potential energy function is
V(R) = - G m1 m2 / R.
Differentiating with respect to R:
dV/dR = - d/dR (G m1 m2 / R) = - (- G m1 m2 / R2) = G m1 m2 / R2.
Thus
F(R) = - dV/dR = - G m1 m2 / R2,
which is Newton’s law (negative sign indicating attraction).
Because potential energy is defined only up to an additive constant, adding any constant C (with dimensions of energy) to V does not change the force, since dC/dR = 0.
So the most general potential consistent with Newton’s law is
V(R) = - G m1 m2 / R + C.
Substituting m1 m2 = 1024:
V(R) = - G / R × 1024 + C.
Detailed Evaluation of Each Option
Option (A): V = - (G / R) × 1024
This matches V(R) = - G m1 m2 / R directly, with m1 m2 = 1024.
Differentiation gives
F(R) = - dV/dR = - (- 1024 G / R2) = - G × 1024 / R2,
which is the correct attractive gravitational force for these masses.
Conclusion for (A): Correct.
Option (B): V = - (G / R) × 1024 + 1000 V0
The first term is the same correct gravitational potential energy as in (A).
The second term, 1000 V0, is a pure constant with dimensions of energy, independent of R.
Since the force is F(R) = - dV/dR, the derivative of a constant is zero, so the force remains
F(R) = - d/dR (- G / R × 1024) = - G × 1024 / R2.
Physics allows such an arbitrary additive constant because only differences in potential energy affect motion.
Conclusion for (B): Also correct.
Option (C): V = G R2 × 1024
Differentiate:
dV/dR = d/dR (1024 G R-2) = 1024 G (-2) R-3 = -2 G × 1024 / R3.
Thus
F(R) = - dV/dR = 2 G × 1024 / R3.
This force varies as 1/R3, not 1/R2, and its magnitude has an extra factor of 2, so it does not match Newton’s law of gravitation.
Conclusion for (C): Incorrect; gives the wrong radial dependence.
Option (D): V = 1012 G R
Differentiate:
dV/dR = 1012 G,
Then
F(R) = - dV/dR = - 1012 G,
a constant attractive force, independent of distance.
Newton’s gravitational force must vary as 1/R2, so this expression cannot represent gravitational potential energy for Newtonian gravity.
Conclusion for (D): Incorrect; predicts a constant force, not an inverse-square law.
Final Answer and Exam-Style Takeaway
The gravitational potential energy between two bodies of masses m1 and m2 separated by distance R is
V(R) = - G m1 m2 / R + constant,
which leads to Newton’s inverse-square gravitational force.
For m1 = m2 = 1012 kg, valid forms are
V(R) = - (G / R) × 1024 and V(R) = - (G / R) × 1024 + 1000 V0.
Hence, the correct options are (A) and (B).
