Question

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

Answer 1

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.)

Answer 2

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