Question

Find the depth at which the value of acceleration due to gravity is same as that a height of 25km

Answer 1

Answer:

Explanation:

The radius of Earth = 6378000 m. Above that, at distance r from Earth’s center, Newton tells us gravity = GM/r^2. Allow me to normalize that to g at 6378000 m:

g_above = g/ (r/6378000)^2 = (6378000/r)^2*g

At 25 km above the surface (note scale cancels)

g_above = (6378/(6378+25))^2*g = 0.9922g

Below the Earth’s surface, gravity drops because the gravitational attraction of mass above radius r cancels out, lessening the mass below by cube of radius, while the inverse square of the shrinking r increases gravity. It turns out that details depend on Earth’s density distribution, but we shall (inaccurately) assume uniform density, so we can use Newton’s other formula (again normalized):

g_below = (r/6378000)g

So we equate g_above for 25 km and g_below, solving for the latter’s radius:

(r/6378000)g = 0.9922g

r = 0.9922*6378000 = 6328.3 km

That is 6378–6328.3 = 49.7 km depth