At what height above the Earth’s surface would the Earth’s gravitational field strength be equal to 7.5 N/kg?

Answer given: 915.8 km

2 Answers
Dwight
Dec 18, 2016

This occurs at a distance of #7.3xx10^6# m from the centre of the Earth (which is 900 km from the surface).

Explanation:

Are we okay to start from the field equation for the Earth's gravitational field?

#g=(GM)/r^2# where M is the mass of the Earth.

What you do now, is change the subject of this equation so it solves for r:

#r = sqrt((GM)/g)#

Plug in the numbers, and you have it!

#r = sqrt(((6.67xx10^(-11))(5.98xx10^(24)))/7.5)#

= #7.3 xx 10^6# m

(By the way, since this value is from the centre of the Earth, and the radius of the Earth is #6.4 xx 10^6# m, we are looking at a point only 900 000 m (900 km) above the surface of the Earth.)

Cesareo R.
Dec 18, 2016

See below.

Explanation:

Given

#G = 6.673 10^(-11)#
#M_(earth) = 5.98 10^24#
#R_(earth) = 6.38 10^6# All in SI units.

We need the value of #r# such that

#(G M_(earth))/(R_(earth)+r)^2=7.5#

Solving for #r# we get

#r = 914.2# Km

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