# A satellite moves in a circular orbit around the earth at a speed of 6.8 km/s. determine the satellites altitude above the surface. assume the earth is a homogeneous sphere of radius 6370 km and mass 5.98 X 10^24 kg. The value of the Iniversal gravitational constant is 6.67 X 10^-11 N X m^2/ Kg^2. Answer in units of km. The gravitational force of attraction between two bodies of mass M1 and M2 and separated by a distance r is given by:

F = G*M1*M2 / r^2

The satellite is revolving around the Earth with a velocity equal to v and at a height of r; the centripetal force is...

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The gravitational force of attraction between two bodies of mass M1 and M2 and separated by a distance r is given by:

F = G*M1*M2 / r^2

The satellite is revolving around the Earth with a velocity equal to v and at a height of r; the centripetal force is equal to mv^2 / r.

As the height of the satellite above the Earth has been taken to be r and its speed is 6.8 km/s, this gives us:

G*Me*Ms / (r + 6370000)^2 = Ms * (6800)^2/ (r + 6370000)

( the distances have been converted to m and r + 6370000 is the distance from the center of the Earth which is required, Ms is the mass of the satellite and gets cancelled)

=> r + 6370000 = G* Me / (6800)^2

=> r = [6.67*10^-11 * 5.98 * 10^24 / (6800)^2] – 6370000

=> r = [6.67*10^-11 * 5.98 * 10^24 / (6800)^2] - 6370000

=> r = 2255994.8 m

=> r = 2255.99 km

The height of the satellite is 2255.99 km

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