A rocket moving around the earth at height "H", If the gravitational acceleration "g1" at height His of gravitational acceleration 'g at earth surface. If Earth radius is "R", find "H"
using R

At the earth's surface  g = G M / R^2        where G is the gravitational constant

at H       g1 = G M / (R + H)^2         using Gauss' theorem for enclosed mass

g1 = G M / (R^2 + 2 R H)    ignoring H^2 as it is small compared to R^2

g / g1 = (R^2 + 2 R H) / R^2 = 1 + 2 R H

g = g1 + 2 R H g1

g1 - g = - 2 R H     or H = (g1 - g) / 2 R

the answer is a. the yacht traveled from the dock north to the 200-meter marker in the bay in less than 5 minutes, giving the passengers several more hours to fish.

im doing peak rn so sry if its wrong

prndulum needs thread, weight clock to time swings