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An astronaut weighted 75 kg on earth. He went on another planet X and weighted himself 50 kg. If the planet X had the same radius as that of earth, then what is the mass of planet with respect to earth ? (M is mass of Earth )
1. 0.5 M
2. 0. 67 M
3. 0. 25 M
4. 1.5 M

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Correct Answer - Option 2 : 0. 67 M


Mass and Weight



Mass is the amount or measure of matter present in a body. Every matter has a mass.

It is a measure of the Normal force we apply to any object. In normal conditions, it is equal to the force of gravity acting on our body.

W = mg

m is mass, g is the acceleration due to gravity

It is fixed for a matter and does not change with the place.

It varies from place to place. It is different on different planets as they have different values of g.

It is having SI unit of Kg. Other units are gram, pound, etc.

It is having SI unit of Newton. Usually, it is also denoted as Kg Wt

A mass having a mass 60 kg will weigh 60 Kg wt. on the earth's surface.

Acceleration Due to Gravity:

  • The force of attraction exerted by the earth on a body is called gravitational pull or gravity.
  • We know that when a force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate.
  • The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity, it is denoted by g.
  • The acceleration due to gravity on the surface is given as

\(g = \frac{{GM}}{{{R^2}}} \)        ---(1)

G is universal constant, M is mass, R is radius of earth.


Given Weight of astronaut on earth W1 = mg = 75 kg wt. 

m is the mass of the object, g is the acceleration due to gravity on earth, 

g' is the acceleration due to gravity on the other planet X

Given Weight of astronaut on Planet X mg' = 50  W2 = kg wt. 

\(\frac{W_2}{W_1} = \frac{mg'}{mg} = \frac{50}{75}\)

⇒ \( \frac{W_2}{W_1} =\frac{g'}{g} = \frac{2}{3}\)

⇒ \(g' = \frac{2}{3} \ g\)        ---(2)

Given the radius of both planet is the same R, 

Mass of earth is M, the mass of planet X is M'

The acceleration due to gravity of another planet is 

\(g' = \frac{{GM'}}{{{R^2}}} \)        ---(3)

Using (1) and (3) in (2) we get

⇒ \(\frac{GM'}{R^2} = \frac{2}{3} \frac{GM}{R^2} \)

⇒ \(M' = \frac{2}{3} M\)

⇒ M' = 0. 67 M

So, the correct option is 0. 67 M.

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