see below...*** on!?

Question

A person weighs 80N on the surface of the earth.What should be the distance of the person from the centre of the earth such that his weight is 5N?
( radius of earth=1600KM)

Answer 1

80N? damn skinny person!

Anyway...

F = GMm/(R^2)
where M=Earth's mass, m = person's mass R= radius of Earth (it's actually more like 6400km) F=80

Solve for m, and then put back into the same equation with F=5 and solve for R.
m=F1*R^2/GM

F2=GMF1*R^2/(GM*r^2)
5/80 = R^2/r^2
r^2 = 16 R^2
r = 4R


Or...there's the simple way of noting that gravity goes with 1/r^2 and that if the force is 1/16th , then the radius must be 4 times as large.

Answer 2

Weight = m g where g = G m M / R^2
where M is mass of earht and R dadius of earth and m mass of a person.

Weight = (80N)=mg..............
m = 80 N / g;
let say at some distance r from contre of earth..
(5N) =(80N/g) * g' where g' is gravity at distance r
sovling this

(r/R)^2 = 16
for r we get the distance as the 4 times the rediu s of earth
So distance is 6400 kms..

Answer 3

w/W = (R/r)^2
r^2=(WR^2)/w
=(80*1600^2)/5 = 16*1600^2
r = 4*1600

r = 6400 km

only, the Earth's equatorial radius, or semi-major axis, is the distance from its center to the equator and equals 6,378.135 km (≈3,963.189 mi; ≈3,443.917 nmi), so

r = 4*6378 = 25,512 km