Question

a man weights 800n at the earth's surface. What will be the weight of the same person when he is orbiting the earth at double the high of the earth's center

Answer 1

200 N

Step-by-step explanation:

F = G m1 m2 / r^2 G m1 and m2 are constant for this scenario

so this becomes

F is proprtional to 1/r^2

f = 1/r^2

800 = 1/r^2 now change r to 2r

F = 1/(2r)^2 = 1/4 * 1/r^2 <======shows force will be 1/4 of original , or 200N