Question

if the earth shrinks to half of its radius,its mass remaining the same,the weight of an object on the earth will become how much times

Answer 1

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

The weight of one 1 kg of body on the Earth is 9.8 N

=> F= (6.67×10⁻¹¹Nm²kg⁻²×6×10²⁴kg×1 kg )/(6.4×10⁶ m)² = 9.8 N

( Mass of the Earth= 6×10²⁴ kg, Radius=6.4×10⁶ m, G= 6.67×10⁻¹¹Nm²kg⁻²)

Now, the radius is (6.4×10⁶)/2= 3.2×10⁶ m, while mass stays the same.

Then the weight of 1 kg object on the Earth will become:

=> (6.67×10⁻¹¹Nm²kg⁻²×6×10²⁴kg×1 kg)/3.2×10⁶ m

=>19.6 N = (9.8 × 2)N

Weight is variable, and since the Earth isn't a perfect sphere, the value of g( acceleration due to gravity) may vary at different places.  

It will become double the weight it was on the Old Earth.

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