Question

1. At what height above the earth is the free-fall acceleration 30% of its value at the surface?


2. What is the speed of a satellite orbiting at that height?

Answer 1

1.
g =GM/R²
g₁=GM/(R+h)²
g₁/g=30/100= GM R²/(R+h)²GM =>
0.3= R²/(R+h)²
Solve for “h” (Earth’s radius is R = 6.378•10⁶ m)
2.
mv²/(R+h)= GmM/(R+h)²
v=sqrt { GM/(R+h)}