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For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is -
A
$\dfrac{1}{\sqrt{2}}$
B
2
C
$\sqrt{2}$
D
$\dfrac{1}{2}$
Detailed Solution
For a satellite of mass $m$ in an orbit of radius $r$: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow mv^2 = \dfrac{GMm}{r}$
Kinetic energy: $K = \dfrac{1}{2}mv^2 = \dfrac{GMm}{2r}$
Potential energy: $U = -\dfrac{GMm}{r}$
$\dfrac{K}{U} = \dfrac{GMm/2r}{-GMm/r} = -\dfrac{1}{2}$
So the magnitude of the ratio of kinetic energy to potential energy is $\dfrac{1}{2}$ (the negative sign only shows that the potential energy is negative).
Kinetic energy: $K = \dfrac{1}{2}mv^2 = \dfrac{GMm}{2r}$
Potential energy: $U = -\dfrac{GMm}{r}$
$\dfrac{K}{U} = \dfrac{GMm/2r}{-GMm/r} = -\dfrac{1}{2}$
So the magnitude of the ratio of kinetic energy to potential energy is $\dfrac{1}{2}$ (the negative sign only shows that the potential energy is negative).