Question

When \(g\) is the acceleration due to gravity on earth, the gain in potential energy of an object of mass \(m\) raised from the surface of earth to a height equal to the radius of earth \(R\) is

• A. \(\dfrac{mgR}{2}\)
• B. \(\dfrac{mgR}{4}\)
• C. \(mgR\)
• D. \(2mgR\)

Hint:

For raising a body from the earth's surface to a height equal to the earth's radius, \[ \Delta U = GMm\left(\frac1R-\frac1{2R}\right) = \frac{GMm}{2R} = \frac{mgR}{2}. \] Do not use \(\Delta U=mgh\) here because \(g\) is not constant over such a large height.

Answer 1

The Correct Option is A

Concept: \(\Delta U = U_f - U_i = -\frac{GMm}{2R} + \frac{GMm}{R} = \frac{GMm}{2R}\). Use \(g = GM/R^2\): \(\Delta U = \frac{mgR}{2}\).

Step 1: Write the final answer. \(\boxed{\Delta U=\frac{mgR}{2}}\)