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

To find the weight of the body at a height above the Earth that is nine times the Earth's radius, we need to understand the concept of gravitational force and how weight changes with distance from the Earth's center. The weight of a body at the Earth's surface is given by \(W = mg\), where \(g\) is the acceleration due to gravity.

According to the law of universal gravitation, the gravitational force \(F\) between two masses is inversely proportional to the square of the distance between their centers. Mathematically, it is expressed as:

\(F = \frac{G M m}{R^2}\)

where:

• \(G\) is the gravitational constant,
• \(M\) is the mass of the Earth,
• \(m\) is the mass of the object,
• \(R\) is the distance from the center of the Earth to the object.

Initially, at the Earth's surface, \(R = R_E\) (radius of the Earth), and the weight \(W = \frac{G M m}{R_E^2}\).

At a height where the body is nine times the radius of the Earth above the surface, the total distance from the Earth's center becomes \(R = 10R_E\) (since \(R_E + 9R_E = 10R_E\)).

The new weight \(W'\) at this height would be:

\(W' = \frac{G M m}{(10R_E)^2} = \frac{G M m}{100R_E^2}\)

Comparing this with the initial weight \(W\):

\(\frac{W'}{W} = \frac{\frac{G M m}{100R_E^2}}{\frac{G M m}{R_E^2}} = \frac{1}{100}\)

Thus, \(W' = \frac{W}{100}\).

Therefore, the correct answer is \(\frac{W}{100}\).