If the earth shrank until its radius were only one-half its present size without changing its mass what would?

Question

a 20N object weigh at its surface?
a.) 2.5 N b) 80N c)100N d) 10N

Answer 1

Formula for gravity = G m1 m2 / r²

So if r is halved then gravity will increase by a factor 4.

==> 80 N

Answer 2

c)80 N.


There is an intuitive way. Since the weight is inversely proportional to the square of the radius, if the radius was half of its original value, the weight should be the reciporacle(excuse spelling) of the square of one half, times its original value. So 20N * ((1/2)^2)^(-1)
Look up these formulas in a text book.

F=Gm1m2/(r^2)
m1 is the mass of the earth. m2 is going to equal the mass that weighs 20N at the surface of the earth so, F=mg
20N=m*9.8
m=20kg/9.8

F=(6.67*10^-11)(5.98*10^24kg)(20/9.8 kg)/((6.37*10^6m*(1/2))^2)
I know that looks like a lot.

Answer 3

the respond is easy as hell in simple terms it s the conservation of angular momentum i.e. mass.velocity.radius So because of the fact that mass is conserved and radius will become r/2 subsequently the linked fee doubles and for this reason the era of an afternoon maintains to be 24 hours :)

Answer 4

Ans- 80N (This is the force applied by earth on the object. Not the weight)

Answer 5

Greetings!

In this particular example the earth would explode.

Good Luck

Answer 6

b) 80N.