<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>
Answer:
Explanation:
The relation between activity and number of radioactive atom in the sample is as follows
dN / dt = λ N where λ is disintegration constant and N is number of radioactive atoms
For the beginning period
dN₀ / dt = λ N₀
58.2 = λ N₀
similarly
41 = λ N
dividing
58.2 / 41 = N₀ / N
N = N₀ x .70446
formula of radioactive decay


- λ t = ln .70446 = - .35
t = .35 / λ
λ = .693 / half life
= .693 / 5715
= .00012126
t = .35 / .00012126
= 2886.36
= 2900 years ( rounding it in two significant figures )
This suggest the the atom has a very small positively charged nucleus in the mass of the atom is concentrated.
And the electrons revolves around the nucleus in their orbits.