PART a. In this orbit, find the force of gravity exerted by the earth on the satellite.
PART b. Find the speed, v, and the period, T, of the satellite.
PART c. During one cycle the satellite is directly above Mt Kenyatta on the equator. Exactly one period later, the satellite is directly above Mt Jomo on the equator. Find the distance on land from Mt Kenyatta to Mt Jomo.
PART d. If the longitude of Mt Kenyatta is 5 degrees West, find the longitude of Mt Jomo.
Known parameters are the earth's mass M= 5.98 E 24 kg, and its radius, R= 6.37 E 6 m, and the constant G= 6.67 E -11 nt m^2/kg^2.
The force of gravity is expressed by:
(1) F= G M m/r^2 in which the orbit radius r = R + h = 6.69 E 6 m. Substituting known quantities should give you gravity force F= 42800 nt.
PART b. The only force on the satellite is the (centripetal) force of gravity toward the c.m. of the earth. Recall from circular motion description that for centripetal acceleration ac= v^2/r, and Newton's second law applied to ...
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