Question

At what height above the earth surface the value of g is half of its value on the earth surface .give its radius is 6000per/km

Answer 1

Answer:

The height above the earth's surface = 2637 km

Explanation:

According to the law of Gravitation,

$F = \frac{GmM}{r^2}$

Here,

• F = Force exerted on the bodies
• G = Gravitational constant
• m = Mass of the body
• M = Mass of the Earth
• r = distance between the body and the core of the Earth.

Acceleration due to gravity,

$g = \frac{F}{m} = \frac{GM}{r^2}$

At surface of Earth,

• r = R

• g = $\frac{GM}{R^2}$

Let the body be at a distance of h from the surface of the Earth and R the radius of the Earth .

• r = R + h

• $g' = \frac{GM}{(R+h)^2}$

According to the question,

At height h,

$g' = \frac{g}{2}$

⇒ $\frac{GM}{R^2} = \frac{GM}{(R+h)^2}$

⇒ 2R² = (R+h)²

⇒ h = (√2 - 1)R

R = 6371 km

∴ h = 2637 km

Thus,

At a height of approximately 2637 km will the acceleration due to gravity will be half of the acceleration due to gravity on the surface