What path will the body cover in the first second if it falls freely from a height of 1000 km?

What path will the body cover in the first second if it falls freely from a height of 1000 km? The radius of the Earth is taken equal to 6400 km.

Free fall acceleration near the Earth’s surface:

g = GM / R ^ 2, where G is the gravitational constant, M is the mass of the Earth, R is the radius of the Earth.

Acceleration at 1000 m:

gh = GM / (R + 1000) ^ 2;

Acceleration ratio:

g / gh = (GM / R ^ 2) / (GM / (R + 1000) ^ 2;

g * GM / (R + 1000) ^ 2 = gh * (GM / R ^ 2);

gh = (g0 * GM / (R + 1000) ^ 2) / (GM / R ^ 2) =
= (g0 * R ^ 2) / (R + 1000) ^ 2 = 0.748g0 =
= 0.748 * 9.8 m / s2 = 7.33 m / s2;

S = ght ^ 2/2 = (7.33 m / s2 * 1 s2) / 2 = 3.67 m.

Answer: 3.67 m.