A man weighs 80 kg on earth's surface. The height above ground where he will weigh 40 kg, is: (Radius of earth is 6400 km)

Question

A man weighs 80 kg on earth's surface. The height above ground where he will weigh 40 kg, is: (Radius of earth is 6400 km) (A) 0.31 times r (B) 0.41 times r (C) 0.51 times r (D) 0.61 times r

Tardigrade Admin · Accepted Answer

Value of acceleration due to gravity decreases on going above the earth's surface.
The weight of a body is given by

w = m g

(on earth's surface)
At a height

h

above the earth's surface

w^{'} = \frac{m g}{( 1 + \frac{h}{R} ) ^{2}}

∴ \frac{w}{w ^{'}} = (1 + \frac{h}{R})^{2}

Given,

w = 80 k g, w^{'} = 40 k g

∴ \frac{80}{40} = (1 + \frac{h}{R})^{2}

\Rightarrow 2 = (1 + \frac{h}{R})^{2}

\Rightarrow 2 = 1.41 = 1 + \frac{h}{R}

\Rightarrow h = 0.41 R

Given,

h = 0.41 r

Q. A man weighs 80kg on earth's surface. The height above ground where he will weigh 40kg, is : (Radius of earth is 6400km)

0.31 times r

0.41 times r

0.51 times r

0.61 times r

Q. A man weighs $80 k g$ on earth's surface. The height above ground where he will weigh $40 k g$ , is : (Radius of earth is $6400 km$ )