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The International Space Station is in a 300-mile-high orbit.Part AWhat is the station's orbital speed? The radius of Earth is 6.37×106m, its mass is 5.98×1024kg.Express your answer with the appropriate units.Part BWhat is the station's orbital period?Express your answer in minutes.

Sagot :

Part A

We will use the next formula to calculate the speed

[tex]v=\sqrt[]{\frac{G\cdot M}{R}}[/tex]

where v is the speed, G is the gravitation constant, M is the mass and r is the radius

First, we need to convert the miles to meters

1 mile =1609m

Height of the orbit = 300 miles= 482700 m

The radius of the Earth = 6.37x10^6m

The radius of the Orbit, R = 6.37x10^6+482700=6852700m

M=5.98x10^24 kg

G= 6.673 × 10-11 N.m^2 / kg^2

Then we substitute

[tex]v=\sqrt[]{\frac{(6.673\times10^{-11})(5.98\times10^{24})}{6852700}}=7630.98\text{ m/s}[/tex]

Part B

For the period T

[tex]T=\frac{2\pi R}{v}[/tex]

We substitute the values

[tex]T=\frac{2\pi(6852700)}{7630.98\text{ }}=5642.4s[/tex]

ANSWER

Part A

7630.98 m/s

Part B

5642.4 s

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