NEET-XI-Physics
exam-2 year:2016
- #124A 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 :-
(1)``\frac{2mg_0R^2}{R + h}``
(2) ``-\frac{2mg_0R^2}{R + h}``
(3) ``\frac{mg_0R^2}{2(R + h)}``
(4) ``-\frac{mg_0R^2}{2(R + h)}``digAnsr: 4Ans : (4)
Sol. Total energy = e
GM m
2(R h)
+
2
e 0
0 e2
GM g R
g M
GR
= &implies; =
∴ Energy =
2
0mg R
2(R h)
+