The potential energy of a satellite of mass m and revolving at a height Rc above the surface of earth where Rc= radius of earth, is

Question

The potential energy of a satellite of mass m and revolving at a height Rc above the surface of earth where Rc= radius of earth, is (A) -mgRc (B) (-m g Rc/2) (C) (-m g Rc/3) (D) (-m g Rc/4)

Tardigrade Admin · Accepted Answer

At a height h above the surface of earth the gravitational potential energy of the particle of mass m is Uh=-(G Me m/Rc+h) Where Mc Rc are the mass radius of earth respectively. In this question, since h=Rc So Uh=Rt=-(G Mc m/2 Rc)=(-m g Rc/2)

Q. The potential energy of a satellite of mass $m$ and revolving at a height $R_{c}$ above the surface of earth where $R_{c} =$ radius of earth, is

$- m g R_{c}$

$\frac{- m g R _{c}}{2}$

$\frac{- m g R _{c}}{3}$

$\frac{- m g R _{c}}{4}$

Q. The potential energy of a satellite of mass m and revolving at a height Rc​ above the surface of earth where Rc​= radius of earth, is

−mgRc​

2−mgRc​​

3−mgRc​​

4−mgRc​​

At a height h above the surface of earth the gravitational potential energy of the particle of mass m is Uh​=−Rc​+hGMe​m​ Where Mc​Rc​ are the mass & radius of earth respectively. In this question, since h=Rc​ So Uh=Rt​​=−2Rc​GMc​m​=2−mgRc​​

Q. The potential energy of a satellite of mass $m$ and revolving at a height $R_{c}$ above the surface of earth where $R_{c} =$ radius of earth, is

$- m g R_{c}$

$\frac{- m g R _{c}}{2}$

$\frac{- m g R _{c}}{3}$

$\frac{- m g R _{c}}{4}$