Question

the earth is shrinks
to 50% of its real radius its mass remain the same the weight of a body on the earth will ​

Answer 1

Answer:The force due to gravity is

m1 is the mass of the Earth and m2 to be the mass of the body.

Force can also be represented as

Further on substitution, the equation becomes

Here we can see that the acceleration due to gravity or g is inversely proportional to Radius or r square.

If mass is kept constant and G being universal gravity constant, radius is halved, then as radius decreases acceleration due to gravity increases, the new acceleration due to gravity becomes 4 times g or 4g.

Weight is calculated as a Force.

Weight = Mass of a body x Acceleration due to gravity.

And hence the weight directly proportional to acceleration due to gravity increases 4 times the original value.

This value is an approximate because here we don't consider the shape of the Earth, Variable Gravity fields and many other external influencing forces.

Answer 2

Earth shrinks 50% of its radius

i.e. becomes Half of the radius now ,

r=R/2------(1)

Mass is same ,

Weight of a body on earth with radius R

i.e. Normal weight.(W¹)

GMm /R² ------(2)

here ,

• G is universal gravitational constant
• M is mass of earth
• m is mass of body
• R is radius of normal earth

Weight of a body on earth with radius r(W²)

GMm /r²------(3)

here ,

• G is universal gravitational constant
• M is mass of earth
• m is mass of body
• r is radius of earth

From eqn (3) and (1)

Weight of object on earth with radius r

=>>> GMm /(R/2)²

>>GMm /(R²/4)

>>4 × GMm/R² -----(4)

From eqn (2) and (4) ,,,

Weight of object on earth with radius r = 4 × Weight of a body on earth with radius R

i.e.

W²=4×W¹

Hope it helps