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Question

A satellite moves around the earth in a circular orbit with speed $$v$$. If $$m$$ is the mass of the satellite, its total energy is


A
12mv2
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B
12mv2
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C
32mv2
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D
14mv2
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Solution

The correct option is A $$-\cfrac { 1 }{ 2 } { m }{ v }^{ 2 }$$
$$T.E= K.E + P.E=    \dfrac{1}{2}mv^2  +  (-\dfrac{GMm}{r})$$ 
Now as   $$F_{centrifugal}= F_{gravitational}     \implies    \dfrac{mv^2}{r}=  \dfrac{GMm}{r^2}$$
Thus,  $$\dfrac{GMm}{r}= mv^2$$
Now,  $$T.E= \dfrac{1}{2}mv^2 - mv^2=  - \dfrac{1}{2}mv^2$$

Physics

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