Gravitation

JEE Main level, single correct, medium

180 of 223

Question 180 of 223

Two satellites AA and BB having masses in the ratio 4:34:3 are revolving in circular orbits of radii 3r3r and 4r4r respectively around the earth. The ratio of total mechanical energy of AA to BB is

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Answer: (B) 16:916:9

Given that m1m2=43,r1r2=34\dfrac{{m}_{1}}{{m}_{2}}=\dfrac{4}{3},\dfrac{{r}_{1}}{{r}_{2}}=\dfrac{3}{4}
Now the total energy of a satellite TE=12mvo2+(−GMmr)TE=\dfrac{1}{2}m{{v}_{o}}^{2}+\left(\dfrac{-GMm}{r}\right)
Since the orbital speed of a satellite vo=GMr{v}_{o}=\sqrt{\dfrac{GM}{r}}
⇒TE=−GMm2r∝mr\Rightarrow TE=-\dfrac{GMm}{2r}\propto \dfrac{m}{r}
⇒TE1TE2=m1m2⋅r2r1=43×43=169\Rightarrow \dfrac{T{E}_{1}}{T{E}_{2}}=\dfrac{{m}_{1}}{{m}_{2}}\cdot \dfrac{{r}_{2}}{{r}_{1}}=\dfrac{4}{3}\times \dfrac{4}{3}=\dfrac{16}{9}

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