Question

A body weight W, is projected vertically upwards from earth's surface to reach a height above the earth which is equal to nine times the radius of earth The weight of the body at that height will be :

• A. $\frac{ W }{100}$
• B. $\frac{ W }{3}$
• C. $\frac{W}{91}$
• D. $\frac{W}{9}$

Hint:

The weight of an object decreases with the square of the distance from the center of the Earth.

Answer 1

The Correct Option is A

\(g'=\frac{GM}{(10R)^2}=(\frac{g}{100})\)
\(W'=(\frac{W}{100})\)
So ,the correct answer is (A) : \(\frac{W}{100}\)

Answer 2

The height of the body at the surface of the Earth = R

The new height of the body reached when projected vertically upward = h = 9R

The weight of the body at the surface of the earth: 
\(W = m\times g \)

The new weight of the body at (h): 
\(W’ = m\times g’\)
\(g’=g\times (\frac{R}{R+h})^2\)

\(g’=g\times (\frac{R}{R+9R})^2\)

\(g’=g\times (\frac{R}{10R})^2\)

\(g’=\frac{g}{100}\)
On putting the value of g’ in \(W’ = m\times g’\), we get:

\(W’=\frac{m\times g}{100}\)

where, \(m\times g = W\), hence: 
\(W’=\frac{W}{100}\)
The weight of the body at the new height (h) will be \(\frac{W}{100}\).