A satellite of mass 6000 kg orbits the Earth in a circular orbit of radius of 9.

Question

A satellite of mass 6000 kg orbits the Earth in a circular orbit of radius of 9.0*10^6 m (this is above the Earth's atmosphere).The mass of the Earth is 6.0 * 10^24 kg.
(a) What is the magnitude of the gravitational force on the satellite due to the earth?
F= N
(b)Using the momentum principle, find the speed of the satellite in orbit.
v= m/s
(c) Using the energy principle, find the minimum amount of work needed to move the satellite from this orbit to a location very far away from the Earth.
work= J

Explanation / Answer

Fg = Fc GMm/r² = m4pi²r/T² GM/r² = 4pi²r/T² M = 4pi²(r^3)/(T²G) = 4pi²(6.13*10^6 + 9.24*10^5)^3/((2.48 h * 3600s/h)^2 * 6.67*10^-11) = 2.61*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)(2.61*10^24)/(6.13*10^6)² = 4.63 m/s² = 4.63 N/kg So, Fg=mg=4000 kg * 4.63 N/kg = 18500 N

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