Find the acceleration due to gravity at an altitude of 2.50 x 10^4 km above the earth's surface.?

Question

also, find the orbital velocity of a satellite at this altitude.

Answer 1

At R = h + r; where h = 2.5 X 10^4 km and r = Earth's radius, the acceleration g = G (r/R)^2; where G = 9.81 m/sec^2 on Earth surface (at r) and g is your acceleration at h above the surface. This results because the force of gravity varies inversely as the square of the distances between the centers of mass. Look up Earth's radius r and you can do the math.

The satellite is orbiting at a fixed orbit R from the center of the Earth. So its weight mg = mv^2/R = centrifugal force. Then we can write v^2 = gR or v = sqrt(gR) and both g and R are known. Again you can do the math.

Bottom line, the physics to remember...g varies inversely with the square of the distance between the two masses (e.g., the satellite and Earth). Also, when in a fixed R orbit, the weight of the satellite is equal and opposite to the centrifugal force on the satellite.

Answer 2

earths radius is about 6378km on average so neglecting the moon and sun it will be zbout 6378^2/(6378+25000)^2 *g

roughly g/24.2