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A remote sensing satellite of the earth revolves in a circular orbit at a height of 250 km above the earth surface. what is the orbital speed and period of revolution of the satellite

User Jeandut
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To find the orbital speed of the satellite, we can use the formula:

v = √(GM/r)

where G is the gravitational constant, M is the mass of the Earth, and r is the radius of the satellite's orbit.

First, we need to convert the altitude of the satellite to the radius of its orbit. The altitude of the satellite is 250 km above the surface of the Earth, so the radius of its orbit is:

r = 250 km + 6378 km = 6628 km

where 6378 km is the radius of the Earth.

Now we can plug in the values and solve for the orbital speed:

v = √(GM/r)
v = √[(6.67 × 10^-11 N·m^2/kg^2)(5.97 × 10^24 kg)/(6.628 × 10^6 m)]
v = 7667 m/s

Therefore, the orbital speed of the satellite is 7667 m/s.

To find the period of revolution of the satellite, we can use the formula:

T = 2πr/v

where T is the period of revolution.

Plugging in the values we just found, we get:

T = 2πr/v
T = 2π(6628 km × 1000)/(7667 m/s)
T = 5425 seconds

Therefore, the period of revolution of the satellite is 5425 seconds, or approximately 90.4 minutes.
User Kevan Ahlquist
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