Calculate the acceleration due to gravity at a height of 300 km from the surface of
the Earth. (M = 5.98 x10²4 kg, R = 6400
km).

8.8854 m/s²

Explanation:

Using the relation :

g = GM / (r + R)^2

G = Gravitational constant = 6.67 * 10^-11

M = mass of earth = 5.98 * 10^24 kg

r = distance = 300 km

R = Radius of earth 6400 km

g = (6.67 * 10^-11 * 5.98 * 10^24) / ((6400+300) * 10^3)^2

g = (6.67 * 5.98 * 10^(-11 + 24)) / (6700 * 10^3)^2

g = (39.8866 * 10^13) / 44890000 * 10^6

g = 0.000000888540 * 10^(13 - 6)

g = 0.000000888540 * 10^7

g = 8.8854 m/s²

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft