Question

The potential energy of a satellite of mass m and revolving at a height $R_{e}$ above the surface of earth where $R_{e}$ = radius of earth, is(a) -mg$R_{e}$(b) $\frac{-mgR_{e}}{2}$(c) $\frac{-mgR_{e}}{3}$(d) $\frac{-mgR_{e}}{4}$

Answer 1

the answer is option b

Answer 2

Answer:

b) -mgRc/2

Explanation:

At a height h above the surface of earth the gravitational potential energy of the particle of mass m

Ub=- GMem/Re+h

In the picture.

where Me & Re are the mass & radius of earth respectively. In this question, since h = Re

So Uh=Rc= -GMem/2Re

                = -mgRe/2

So the potential energy of the satellite is Uh=-mRe/2