Question

A satellite orbits the earth at a radius RS = 12.4 RE (measured from the center of the earth), where RE is the radius of the earth (take it to be 4000 mi with there being 5,280 ft/mi). [Hint: recall that the acceleration of a satellite at the surface of the earth where RS =RE is g=32ft/sec^2 and the acceleration varies inversely as the square of the distance from the earth's center.] What is the speed v of the satellite in mi/hr (miles per hour)?

Answer 1

Using Force balance on satellite's circular motion

Centripetal Force = Gravitational force on Satellite

Fc = Fg

m*V^2/R = G*m*Me/R^2

V = sqrt (G*Me/R)

R = 12.4*Re

V = sqrt (G*Me/(12.4*Re))

gravitation field of earth is given by:

g = G*Me/Re^2

G*Me/Re = g*Re

So,

V = sqrt (g*Re/12.4)

Using given values:

g = 32 ft/sec^2

Re = 4000 mi = 4000 mi*(5280 ft/mi) = 21120000 ft

So,

V = sqrt (32*21120000/12.4)

V = 7382.63 ft/sec = (7382.63ft/sec)*(1 mile /5280 ft)*(3600 sec/1 hr) = 5033.6 mi/hr

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