Question

A 650-kg satellite is in a circular orbit about Earth at a height above Earth equal to

Answer 1

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)

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

V=5.59*10^3m/s

b)

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

T=3.98hours

c)

$F=\frac{GMM}{R+h^2}$

Therefore

$F=\frac{6.67*10^{-11}*5.98*10^{24}*650}{2*6.38*10^6}$

F=1.47*10^3N

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