Answer to Question #152440 in Mechanics | Relativity for Ira

Question

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.                                                         

        

Answer 1

1. We have artificial satellite 

Let's write a given constants

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

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

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

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

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

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

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

$T = \frac{2\pi R}{v} = \frac{2\pi R}{\sqrt{aR}} = \frac{2\pi R}{\sqrt{\frac{GM}{R^2}R}} = \frac{2\pi R\sqrt{R}}{\sqrt{GM}} = \frac{6.28 * 7 * 10^7 * \sqrt{7 * 10^7}}{\sqrt{6.67*10^{-11} M}}$ = 1.77 * 86400s

$M = 1.8*10^{27}kg$