Verified Solution • 2025–26 Session moderate

QNEET-2016-Ph-1177: NEET 2016 Physics - Class 11 NEET-2016-Ph-1177 Solution: A satellite of mass m is orbiting the earth (of radius R) at a…

A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of

g_0

the value of acceleration due to gravity at the earth's surface is

\frac{mg_0R^2}{2(R+h)}

-\frac{mg_0R^2}{2(R+h)}

\frac{2mg_0R^2}{R+h}

-\frac{2mg_0R^2}{R+h}

A ${{m{g_0}{R^2}} \over {2\left( {R + h} \right)}}$ 2 ( R + h ) m g 0 _ R 2 _
B $- {{m{g_0}{R^2}} \over {2\left( {R + h} \right)}}$ _ 2 ( R + h ) m g 0 _ R 2 _
C ${{2m{g_0}{R^2}} \over {R + h}}$ R + h 2 m g 0 _ R 2 _
D $- {{2m{g_0}{R^2}} \over {R + h}}$ _ R + h 2 m g 0 _ R 2 _

Step-by-Step Verified Solution

Total mechanical energy of satellite is given by E = � GMm/2r Here, r = R + h And, GM = g o R 2 So, E = � mg o R 2 /2(R+h)