Answer to Question #204150 in Classical Mechanics for Mile

Question

Use the mass of the Earth as 6x1024 kg. A 250 kg satellite is in orbit around the Earth 150 km above the Earth=s surface. (Rearth = 6.38 x 106 )

A) What is the gravitational force felt by the satellite?

B) What is the centripetal acceleration?

C) What is the orbital speed of the satellite?

D) What is the satellites period in s? min? hr?

Answer 1

$m_E = 6*10^{24}kg$

$m_s =250kg$

$h=250km=25*10^4m$

$r_E=6.38*10^6m$

$G =6.67*10^{-11}m^3s^{-2}kg^{-1}$

$A) \ \vec F= G\frac{m_Em_s}{(r_E+h)^2}$

$\vec F= 6.67*10^{-11}*\frac{6*10^{24}*250}{(6.38*10^6+25*10^4)^2}=2276.1$

$\text{Answer:}2276.1N$

$B)\vec F=m\vec a;\vec a =\frac{\vec F}{m}$

$\vec a=\frac{2276.1}{250}=9.1$

$\text{Answer: }\vec a = 9.1\frac{m}{s^2}$

$C) \vec a=\frac{\vec v^2}{r};r = r_E+h$

$\vec v =\sqrt{ (r_E+h)*\vec a}$

$\vec v = \sqrt{(6.38*10^6+25*10^4)*9.1}=7767.4$

$\text{Answer: }\vec v=7767.4 \frac{m}{s}$

$D) s= \vec vT;s= 2*\pi*(r_E+h)$

$T = \frac{2*\pi*(r_E+h)}{\vec v}$

$T = \frac{2*3.14*(6.38*10^6+25*10^4)}{7767.4}=5360$

$T =5360 {sec}= 1h\ 29min\ 20 sec$

$\text{Answer: }T =5360 {sec}= 1h\ 29min\ 20 sec$