Question

5. Given that acceleration due to gravity varies inversely as
the square of the distance from the center of earth, find its
value at a height of 64 km from the earth's surface, if the
value at the surface be 9.81 ms^-2. Radius of earth = 6400
km.​

Answer 1

Given that acceleration due to gravity varies inversely as the square of the distance from the center of earth, find its value at a height of 64 km from the earth's surface, if the value at the surface be 9.81 ms^-2. Radius of earth = 6400km.

acceleration due to gravity at height h from the surface of the earth is given by, $g=\frac{g_0}{\left(1+\frac{h}{R}\right)^2}$

where $g_0$ is acceleration due to gravity at the surface of the earth.

here, $g_0$ = 9.81 m/s²

h = 64km , R = 6400 km

so, g = 9.81/(1 + 64/6400)²

= 9.81/(1 + 1/100)²

= 9.81/(1.01)²

= 9.61670424 ≈ 9.616 m/s²

hence, acceleration due to gravity at a height 64km from the earth's surface is 9.616 m/s²

Answer 2

Answer:

9.62  m/s²

Step-by-step explanation:

Given that acceleration due to gravity varies inversely as

the square of the distance from the center of earth, find its

value at a height of 64 km from the earth's surface, if the

value at the surface be 9.81 ms^-2. Radius of earth = 6400

km.​

acceleration due to gravity ∝  1/(distance from the center of earth)²

acceleration due to gravity = k/(distance from the center of earth)²

k being constant

gs = acceleration due to gravity  at surface

gs = k/(6400)² = 9.81   ( given)

=> k = 9.81 * (6400)²

g₆₄ = acceleration due to gravity  at 64 km height

g₆₄ = k/(6400+64)²

putting value of k

g₆₄ = 9.81 * (6400)² /(6464)²

=> g₆₄ = 9.81/(1.01)²

=> g₆₄ = 9.81/(1.0201)

=> g₆₄ = 9.62 m/s²

acceleration due to gravity  at a height of 64 km from the earth's surface = 9.62  m/s²