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A 650-kg satellite is in a circular orbit about Earth at a height above Earth equal to
Earth’s mean radius. Find (a) the satellite’s orbital speed, (b) the period of its
revolution, and (c) the gravitational force acting on it.

Respuesta :

okpalawalter8

For a 650-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth’s mean radius, the resultant values are is mathematically given as

  • V=5.59*10^3m/s
  • T=3.98hours
  • F=1.47*10^3N

What are the satellite’s orbital speed, the period of its revolution, and the gravitational force acting on it.?

Generally, the equation for satellite orbital speed is mathematically given as

a)

[tex]V^2=\frac{Gm}{R+h}\\\\Therefore\\\\V^2=\frac{6.667*10^-11*5.98*10^{24}}{2*6.38*10^6}[/tex]

V=5.59*10^3m/s

b)

[tex]V=2\pi(R+h)/T\\\\Hence\\\\T=2\pi(R+h)/V\\\\T=2\pi(R+h)/2*3.14*6.38*10^6/5.59*10^3[/tex]

T=3.98hours

c)

[tex]F=\frac{GMM}{R+h^2}[/tex]

Therefore

[tex]F=\frac{6.67*10^{-11}*5.98*10^{24}*650}{2*6.38*10^6}[/tex]

F=1.47*10^3N

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