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The International Space Station (ISS) orbits Earth at an altitude of 400 km. Using this information, plus the mass and radius of Earth, calculate the orbital velocity of the ISS (in m/s) and the period of a single orbit around Earth (in minutes).

User Durga Dutt
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1 Answer

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Answer:

v = 7671.57 m/s

T = 1.55 hours

Step-by-step explanation:

mass of Earth, M = 6 x 10^24 kg

Radius of earth, R = 6400 km = 6.4 x 10^6 m

height, h = 400 km

Velocity is given by


v=\sqrt{(GM)/(R+h)}

where, G be the universal gravitational constant.

G = 6.657 x 10^-11 Nm^2/kg^2


v=\sqrt{(6.67* 10^(-11)* 6* 10^(24))/(6800* 10^(3))}

v = 7671.57 m/s

Let T b the period


T=(2\pi (R+h))/(v)


T=(2* 3.14(6800* 1000))/(7671.57)

T = 5566.53 second

T = 1.55 hours

User ScreechOwl
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