A satellite has a mass of 5750 kg and is in a circular orbit 3.50 105 m above th

Question

A satellite has a mass of 5750 kg and is in a circular orbit 3.50 105 m above the surface of a planet. The period of the orbit is two hours. The radius of the planet is 4.05 106 m. What is the true weight of the satellite when it is at rest on the planet's surface

I used the equation mrw2=GMm/R2 to find M then put M in the eqaution W=GMm/R2

The R in 1st equation = R=+h

I keep failing to get the correct answer, Please point out where my equations in wrong.

Answer is in Newtons and I keep getting 1810 N as my answer.

Explanation / Answer

The centripetal force is provided by the force of gravity.

Fg = Fc
GMm/r² = m4pi²r/T²
GM/r² = 4pi²r/T²
M = 4pi²(r^3)/(T²G)
= 4*3.14*3.14(4.05*10^6 + 3.50*10^5)^3/((2.2 * 3600)^2 * 6.67*10^-11)
= 0.802*10^24 [this is the mass of the planet]

The gravitational force, Fg, is also equal to m*a. To find the acceleration,

ma=GMm/r²
a=GM/r²

At the surface of the planet, a=(6.67*10^-11)(0.802*10^24)/(4.05*10^6)^2 = 3.261 m/s² = 3.261 N/kg

So, Fg=mg=4000 kg * 3.261 N/kg =13044 N

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