Test Index

Gravitation Part 1

Section: Physics

Question : 30 of 100

Marks: +1, -0

Two satellites

A

B

200\ \text{kg}

400\ \text{kg}

600\ \text{km}

1600\ \text{km}

T_{A}

T_{B}

A

B

T_{B}-T_{A}

=6400\ \text{km}

=6 \times 10^{24}\ \text{kg}

Solution:

Given,

M_{A}=200\ \text{kg},\ M_{B}=400\ \text{kg},\ H_{A}=600\ \text{km},\ H_{B}=1600\ \text{km}

and

\;\; R_{A}=R_{E}+H_{A}=6400+600=7000\ \text{km}

R_{B}=R_{E}+H_{B}=6400+1600=8000\ \text{km}

Let

T_{A},\ T_{B},\ \omega_{A},\ \omega_{B},\ R_{A}

and

R_{B}

be the time period, angular frequencies and radii of satellites

A

and

B

, respectively. Force on satellite

A,\ F_{A}=m_{A} \omega_{A}^{2} R_{A}=\;\frac{G M m_{A}}{R_{A}^{2}}

\Rightarrow \;\; \omega_{A}^{2}=\;\frac{G M}{R_{A}^{3}}

but,

\;\; \omega_{A}=\;\frac{2 \pi}{T_{A}}

\therefore \;\; \left(\frac{2 \pi}{T_{A}}\right)^{2}=\;\frac{G M}{R_{A}^{3}} \Rightarrow T_{A}=\sqrt{\;\frac{4 \pi^{2} R_{A}^{3}}{G M}}

Similarly,

\;\; T_{B}=\sqrt{\;\frac{4 \pi^{2} R_{B}^{3}}{G M}}

\therefore \;\; T_{B}-T_{A} \;\; =\sqrt{\;\frac{4 \pi^{2}}{G M}}\left(\sqrt{R_{B}^{3}}-\sqrt{R_{A}^{3}}\right)

\;\; =\;\frac{2 \pi}{\sqrt{G M}}\left[\sqrt{(8 \times 10^{3})^{3}}-\sqrt{(7 \times 10^{3})^{3}}\right]

=\;\frac{2 \pi \times 10^{9}}{\sqrt{G M}}(8 \sqrt{8}-7 \sqrt{7})

=\;\frac{2 \pi \times 10^{9}}{\sqrt{6.67 \times 10^{-11} \times 6 \times 10^{24}}}(4.107)

=\;\frac{2 \pi}{\sqrt{4 \times 10^{14}}} \times 10^{9}(4.107)

=\pi \times 10^{2} \times 4.107=12.9 \times 10^{2}\ \text{s}

=1.33 \times 10^{3}\ \text{s}

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