Question #152440

1.     An artificial satellite circles the earth in a circular orbit at a location where the acceleration due to gravity is 9.0ms-2. Determine the orbital period of the satellite.

2.  A satellite of Jupiter has an orbital period of 1.77 days and an orbital radius of . Determine the mass of Jupiter.                                                         

        


1
Expert's answer
2020-12-23T07:32:45-0500
  1. We have artificial satellite 

Let's write a given constants

G = 6.67 * 10-11 [m3s2kg\frac{m^3*s^2}{kg}] and MearthM_{earth} = 6 * 1024 kg a = 9[ms2\frac{m}{s^2}]

First, we compute R from a = GMR2\frac{GM}{R^2}, We find R = GMa\sqrt{\frac{GM}{a}}

Second we compute v from a = v2R\frac{v^2}{R} , which gives v = aR\sqrt{aR} = GMa4\sqrt[4]{GMa}

Finally, we compute T from this v = 2πRT\frac{2\pi R}{T}

T = 2πRv=2πGMa1GMa4=2πGMa34\frac{2\pi R}{v} = 2\pi * \sqrt{GMa} * \frac{1}{\sqrt[4]{GMa}} = 2\pi * \sqrt[4]{\frac{GM}{a^3}}

= 6.28 * 6.67101161024274\sqrt[4]{\frac{6.67 * 10^{-11} * 6 * 10^{24}}{27}} = 12322[s] = 3.42[hour]

  1. We have T = 1.77days = 1.77 * 86400s and orbital radius R = 7 * 107 m

T=2πRv=2πRaR=2πRGMR2R=2πRRGM=6.28710771076.671011MT = \frac{2\pi R}{v} = \frac{2\pi R}{\sqrt{aR}} = \frac{2\pi R}{\sqrt{\frac{GM}{R^2}R}} = \frac{2\pi R\sqrt{R}}{\sqrt{GM}} = \frac{6.28 * 7 * 10^7 * \sqrt{7 * 10^7}}{\sqrt{6.67*10^{-11} M}} = 1.77 * 86400s

M=1.81027kgM = 1.8*10^{27}kg


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