Question

Two satellites are in orbit around a planet. One satellite has an orbital radius of 8.0×1 0 6 m. The period of revolution for this satellite is 1.0×1 0 6 s. The other satellite has an orbital radius of 2.0×1 0 7 m. What is this satellite’s period of revolution?

Answer 1

This question involves the concept of the orbital period.

The period of revolution of the second satellite is "3.95 x 10⁶ s".

ORBITAL PERIOD

First, we will consider the orbital period of the first satellite:

$T_1=\sqrt{\frac{4\pi^2 r_1^3}{GM}}$

where,

• T₁ = orbital period of frirst satellite = 1 x 10⁶ s
• r₁ = orbital radius for first satellite = 8 x 10⁶ m
• M = mass of planet = ?
• G = Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

Therefore,

$1\ x\ 10^6\ s=\sqrt{\frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(M)}}\\\\M = \frac{4\pi^2(8\ x\ 10^6\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(1\ x\ 10^6\ s)^2}\\\\M = 3.03\ x\ 10^{20}\ kg$

Now, we find out the orbital period of the second satellite:

$T_2=\sqrt{\frac{4\pi^2 r_2^3}{GM}}$

where,

• T₂ = orbital period of second satellite = ?
• r₁ = orbital radius for second satellite = 2 x 10⁷ m

Therefore,

$T_2=\sqrt{\frac{4\pi^2(2\ x\ 10^7\ m)^3}{(6.67\ x\ 10^{-11}\ N.m^2/kg^2)(3.03\ x\ 10^{20}\ kg)}}$

T₂ = 3.95 x 10⁶ s

Learn more about the orbital period here:

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