A 3000-kg satellite orbits the Earth in a circular orbit 11797 km above the Earth's surface (Earth radius = 6380 km, Earth Mass = 5.97x10^24 kg). Reminders:Distance should be in meters, not kilometers. 1000 m = 1 km.The total radius needed for the problem is r=r earth + hightWhat is the gravitational force (in newtons, N) between the satellite and the Earth?Hint: The radius of the Earth + the height of the orbit = the center-to-center distance needed for the equation. You also need the universal gravitational constant (G), which is not 9.81 m/s^2. Be careful.Fg=Gm1m2/r2Answer: __________ N

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AriamG79670 AriamG79670 · Answer 1 · 2022-10-25T11:18:53+02:00

We have:

m1 = mass 1 = 3000 kg

h = height = 11797 km = 11797000 m

r2 = 6380 km = 6380000

m2 = mass 2 = 5.97x10^24 kg

G = gravitational constant = 6.6743 × 10-11 Nm^2 /kg^2

r= distance = h + r2 = 11797000 m + 6390000 m = 18,177,000 m

Apply:

Fg = G m1m2/ r^2

Replacing:

Fg = 6.6743 × 10-11 Nm^2/kg^2 ( 3000 kg * 5.97x10^24 kg ) / (18,177,000 m)^2

Fg= 3,617.9 N

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