Gravity Radius of earth and weight

Classical Mechanics Level 2

If R is the radius of the Earth, then the height at which the weight of a body becomes

1/4th of its original weight (as on the surface of the Earth) is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ ?

For more such problems, try my set Gravity

R/4 R R/2 4R 2R

1 solution

Parag Zode
Feb 5, 2015

$g'=\dfrac{gR^{2}}{(R+h)^2}$

$\dfrac{g'}{g} = \dfrac{R^2}{(R+h)^2}$

Since $g'=1$ and $g=4$

$\dfrac{1}{4} = \dfrac{R^2}{(R+h)^2}$

$\dfrac{1}{2} = \dfrac{R}{(R+h)}$

$R+h=2R$

$h=2R-R=\boxed{R}$

you've got it absolutely right! That was cool sir!

Sravanth C. - 6 years, 4 months ago

Thanks @Sravanth Chebrolu . $\ddot\smile$

Parag Zode - 6 years, 4 months ago

Absolutely sir! your welcome. That's my pleasure......

But sir can you please tell me how to derive the formula you've used (only if you have time to spare) I have tried with no success...

I hope you would help...

Sravanth C. - 6 years, 4 months ago

@Sravanth C. – Weight of the object on surface of Earth = Gravitational Force.

= $mg = \dfrac {GMm}{R^2}$

$\implies g = \dfrac {GM}{R^2}.....(1)$

At Height $h$ from the Earth's surface ,acceleration due to gravity is $g'$

At height $h$ ,

Weight of the object = Gravitational Force.

$mg' = \dfrac{GMm}{(R+h)^2}$

$g' = \dfrac{GM}{(R+h)^2}.....(2)$

From equations $(1)$ and $(2)$ ,we get :-

$g' = \ \dfrac {gR^{2}}{(R+h)^2}$

It shows that $g'$ depends upon height of the object above the surface of the earth . It is independent of the mass of object. $\ddot\smile$

Parag Zode - 6 years, 4 months ago

@Parag Zode – thank you very much sir!...

Sravanth C. - 6 years, 4 months ago