Two identical satellites are orbiting at a distance R & 7R from the surface of the earth, R being the radius of the earth. The ratio of their

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T.E’s is 1 : 1

T.E’s is 4 : 1

P.E’s is 4 : 1

K.E’s is 4 : 1

answer is A, B, C.

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Detailed Solution

$(i) h_{1} = R, h_{2} = 7 R$

$K . E_{1} = \frac{G M m}{2 (R + R)}, K_{2} = \frac{G M m}{2 (R + 7 R)}$

$\begin{aligned} \frac{K_{1}}{K_{2}} = \frac{16}{4} = 4 \\ K_{1} : K_{2} = 4 : 1 \end{aligned}$

$(ii) P \cdot E_{1} = \frac{- G M m}{R + R}, P \cdot E_{2} = \frac{- G M m}{R + 7 R}$

$\begin{aligned} \frac{P_{1}}{P_{2}} = \frac{8}{2} \frac{R}{R} = \frac{4}{1} \\ P_{1} : P_{2} = 4 : 1 \end{aligned}$

$\begin{aligned} (iii) T . E = \frac{- G M m}{2 (R + h)} \\ T.E E_{1} = \frac{- G M m}{4 R} \end{aligned}$

$T \cdot E_{2} = \frac{- G M m}{2 (R + 7 R)} = \frac{- G M m}{16 R}$

$\frac{T \cdot E_{1}}{T . E_{2}} = \frac{4}{1}$

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