Question

Assuming that the gravitational potential energy of an object at infinity is zero, the change in potential energy (final - initial) of an object of mass m, when taken to a height h from the surface of earth (of radius R), is given by,

• A. $-\frac{GMm}{R + h}$
• B. $\frac{GMmh}{R\left(R + h\right)}$
• C. mgh
• D. $\frac{GMm}{R + h}$

Answer 1

The Correct Option is B

The question involves calculating the change in gravitational potential energy of an object when moved to a certain height from the surface of the Earth. Let's solve this step-by-step:

1. U = -\frac{GMm}{r}, where U is the gravitational potential energy, G is the gravitational constant, M is the mass of the Earth, m is the mass of the object, and r is the distance from the center of the Earth.
2. Initially, the object is on the surface of the Earth, so the initial potential energy U_i is: U_i = -\frac{GMm}{R} where R is the radius of the Earth.
3. When the object is raised to a height h, the distance from the center of the Earth becomes R + h. Thus, the final potential energy U_f is: U_f = -\frac{GMm}{R + h}
4. The change in potential energy is calculated as: \Delta U = U_f - U_i = \left(-\frac{GMm}{R + h}\right) - \left(-\frac{GMm}{R}\right)
5. Simplifying the above expression gives us: \Delta U = \frac{GMm}{R} - \frac{GMm}{R + h}
6. Taking common denominators, we get: \Delta U = \frac{GMm \left( R + h - R \right)}{R(R + h)} which simplifies to: \Delta U = \frac{GMmh}{R(R + h)}

Thus, the correct formula for the change in gravitational potential energy is \frac{GMmh}{R(R + h)}, which corresponds to the second option.

This derivation assumes the potential energy at infinity is zero, making the potential energy negative and dependent on the distance from the center of the Earth, as per the gravitational potential energy formula.