Assuming that the Earth has a uniform density 𝜌=5540.0 kg/m3, what is the value of the gravitational acceleration at a distance 𝑑=1300.0 km from the Earth's center?

Question

Assuming that the Earth has a uniform density 𝜌=5540.0 kg/m3, what is the value of the gravitational acceleration at a distance 𝑑=1300.0 km from the Earth's center?

Answer 1

The distance of 1300 km happens to be less than the Earth's radius.

So we need to divide the Earth into two parts:

- a sphere of radius d=1300km, and

- the remainder, which is a hollowed out sphere.

Both parts can contribute gravitational acceleration.

The key to solving this problem is the property that gravitational acceleration inside a hollowed out sphere

is 0.

I do not know whether you were just told that, or you are supposed to derive it, but let's take it for given.

So the only cause of gravitational acceleration there is the sphere of radius d=1300 km.

And its gravitational field is the same as if all the mass were in the center.

The formula for gravitational acceleration is

a = GM/d²,

where G is the gravitational constant, and

M is the mass of the sphere.

M = ρV,

where V is the volume of the sphere.

The volume of a sphere of radius d is

V = 4πd³/3

Putting it together

a = (Gρ4πd³/3)/d² = 4Gρπd/3

Substituting actual numbers

a = 4 × 6.6743×10^-11 × 5540 × π × 13×10⁵ / 3 = 2.01 m/s²