Determine the ratio of earth's gravitational force exerted on an 80-kg person when at earth's surface and when 1400 km above earth's surface. the radius of earth is 6370 km. masteringphysics feonp, surfacefeonp, h = 1.49 put 1.49

m = mass of the person = 80 kg

M = mass of earth = 5.98 x 10²⁴ kg

R = radius of earth = 6.37 x 10⁶ m

h = height above the earth's surface = 1400 km = 1.4 x 10⁶ m

r₁ = initial distance of the person from the center of earth when on surface = R =  6.37 x 10⁶ m

r₂ = final distance of the person from the center of earth when at some height = R + h =  6.37 x 10⁶ + 1.4 x 10⁶ = 7.77 x 10⁶ m

F₁ = Gravitational force of earth on the person when at surface

Gravitational force of earth on the person when at surface is given as

F₁ = G M m/r₁²                                             eq-1

F₂ = Gravitational force of earth on the person when at some height

Gravitational force of earth on the person when at some height is given as

F₂ = G M m/r₂²                                             eq-2

dividing eq-1 by eq-2

F₁ /F₂ = (G M m/r₁² )/(G M m/r₂²)

F₁ /F₂ = r₂²/r₁²

inserting the values

F₁ /F₂ = (7.77 x 10⁶)²/(6.37 x 10⁶)²

F₁ /F₂ = 1.49

volume: 22

density: 7.04

elctricity and magnet is conversion to it