A planet has two small satellites in circular orbits around the planet. The first satellite has a period 18.0 hours and an orbital radius 2.

Question

A planet has two small satellites in circular orbits around the planet. The first satellite has a period 18.0 hours and an orbital radius 2.00 × 10 7 m. The second planet has an orbital radius 4.00 × 10 7 m. What is the period of the second satellite? g = 9.80 m/s^2

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Khoii Minh 4 days 2021-07-22T07:52:39+00:00 1 Answers 0 views 0

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    2021-07-22T07:53:46+00:00

    Answer:

    Time period of second satellite will be 50.904 hour

    Explanation:

    We have given time period of first satellite T_1=18hour

    Orbital radius of first satellite a_1=2\times 10^7m

    Orbital radius of second satellite a_2=4\times 10^7m

    We have to find the time period of second satellite.

    Time period of satellite is equal to T=2\pi \sqrt{\frac{a^3}{G(M_1+M_2)}}

    From the relation we can see that T\propto a^{\frac{3}{2}}

    \frac{T_1}{T_2}=\frac{a_1^\frac{3}{2}}{a_2^\frac{3}{2}}

    \frac{18}{T_2}=\frac{(2\times 10^7)^\frac{3}{2}}{(4\times 10^7)^\frac{3}{2}}

    \frac{18}{T_2}=(\frac{1}{2})^\frac{3}{2}

    T_2=50.904hour

    Time period of second satellite will be 50.904 hour

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