Question

The International Space Station orbits at an average height of 300 km above sea level. (a) Determine the acceleration due to gravity at that height and find the orbital velocity and the period of the space station. (Assume the radius and mass of the Earth are 6.37 103 km and 5.97 1024 kg respectively.) orbital velocity m/s acceleration due to gravity m/s2 period s (b) The Hubble Space Telescope orbits at 580 km. What is the telescope's orbital velocity and period? orbital velocity m/s period s

Answer #1

given

height of 300 km = h

radius and mass of the Earth are 6.37 X 10^{3} km =
R

m = 5.97 X 10^{24} kg

using T = 2 r / V

GM / r = (2 r / T
)^{2}

6.67 X 10^{-11} X 5.97 X 10^{24} / ( 6.37 X
10^{6} + 300 X 10^{3} )^{3} = (2 / T
)^{2}

(2 / T )^{2} =
1.341 X 10^{-6}

2 / T = 1.158 X
10^{-3}

**T = 5423.1433 sec**

velocity is V = ( 6.67 X 10^{-11} X 5.97
X10^{24} / ( 6.37 X 10^{6} + 300 X 10^{3} )
)^{1/2}

V = 59700000^{1/2}

**V = 7726.577 m/s**

**the telescope's orbital velocity** **V =
7726.577 m/s**

a = V^{2} / R+h

= 7726.577^{2} / 6.37 X10^{6} + 300 X
10^{3}

**a = 8.95 m/sec ^{2}**

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