Question

If the earth shrinks 50% of its real radius its mass remaining the same the weight of a body on the earth will
Explain this

Answer 1

Let radius of earth is R .

then, acceleration due to gravity on the earth's surface is given by, $g=\frac{GM}{R^2}$

and then, weight of body of mass m, is W = mg

a/c to question, The earth shrinks 50% of its real radius .

e.g., new radius , R' = 50% of R = R/2

then, acceleration due to gravity on the earth's surface is given by, $g'=\frac{GM}{R'^2}$

or, $g'=\frac{GM}{R^2/4}=\frac{4GM}{R^2}= \frac{g}{4}$

then, weight of body of mass m will be W' = mg' = mg/4

hence, weight of body will be (1/4)th of weight of body when radius of earth is real.

Answer 2

If acceleration due to gravity on earth(g) = G.M/R²,

then weight of the body = mg;

where G is gravitation constant, M is mass of earth and R is the radius.

Now if the radius is reduced by half it means g is reduces by 1/4th because it is inversely proportional to the square of radius.

And weight of the body is directly proportional to g, it will also reduce by a factor of 4.