<span>Let F be the force of gravity, G be the gravitational constant, M be the mass of the earth, m your mass and r the radius of the earth, then:
F = G(Mm / (4(pi)*r^2))
The above expression gives the force that you feel on the earth's surface, as it is today!
Let us now double the mass of the earth and decrease its diameter to half its original size.
This is the same as replacing M with 2M and r with r/2.
Now the gravitational force (F' ) on the new earth's surface is given by:
F' = G(2Mm / (4(pi)(r/2)^2)) = 2G(Mm / ((1/4)*4(pi)*r^2)) = 8G(Mm / (4(pi)*r^2)) = 8F
So:
F' = 8F
This implies that the force that you would feel pulling you down (your weight) would increase by 800%!
You would be 8 times heavier on this "new" earth!</span>
S = ut + 1/2 at^2
a = 3.2 m/s^2
s = 15m
Find t
15 = 1/2(3.2)t^2
15 = 3.2t^2/2
30 = 3.2t^2
30/ 3.2 = 9.38
Square root of 9.38 = 3.06
It takes 3.06 seconds
I have to guess it is force
