Practicing Success
Practicing Success
CUET Preparation Today
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 go, the value of acceleration due to gravity at the earth's surface, is : |
\(\frac{mg_oR^2}{2(R+h)}\) - \(\frac{mg_oR^2}{2(R+h)}\) \(\frac{2mg_oR^2}{(R+h)}\) - \(\frac{2mg_oR^2}{(R+h)}\) |
- \(\frac{mg_oR^2}{2(R+h)}\) |
Energy : E = - \(\frac{GMm}{2r}\) r = R + h ; GM = goR2 E = - \(\frac{mg_oR^2}{2(R+h)}\) |
