If a man weighs 60kg on earth surface , the height above the surface of tge earth where his weight is 30 kg is ( assume radius of earth r)
Answers
As the weight is twice on earth surface and mass is constant we would solve it by gravity variation due to height
As the weight is half above some height that means gravity is half their
g'=g(1-2h/r)
g/2=g(1-2h/r)
1/2=1-2h/r
2h/r=1/2
h=r/4
It means r/4(where r is radius of Earth) height above the earth surface the weight would be 30
Answer:
0.41 R
Given:
Mass of man on surface of Earth= 60kg
Mass of man on above the surface of Earth = 30kg
\Large\underline{\underline{\sf To\:Find:}}
ToFind:
Height = ?
\Large\underline{\underline{\sf Formula\:Used:}}
FormulaUsed:
\large{\boxed{\boxed{\sf \pink{F=\dfrac{GMm}{R^2}}}}}
F=
R
2
GMm
\Large\underline{\underline{\sf Solution:}}
Solution:
Let the height is h
Force is inversely proportional to Square of radius.
\Large{\sf \red{F\:\:\:\alpha\:\:\:\dfrac{1}{R^2}}}Fα
R
2
1
⛬
\implies{\sf \dfrac{F_1}{F_2}=\dfrac{(R+h)^2}{R^2}}⟹
F
2
F
1
=
R
2
(R+h)
2
\implies{\sf \dfrac{60}{30}=\left(\dfrac{R+h}{R}\right)^2 }⟹
30
60
=(
R
R+h
)
2
\implies{\sf \sqrt{2}=\dfrac{R+h}{R}}⟹
2
=
R
R+h
\implies{\sf R+h=\sqrt{2}R }⟹R+h=
2
R
$$\implies{\sf h=\sqrt{2}R-R}}$$
$$\implies{\sf h=( \sqrt{2}-1)R }$$
$$\implies{\sf h=(1.41-1)R}$$
$$\implies{\sf \red{h=0.41R }}$$
$$\Large\underline{\underline{\sf Answer:}}$$
Option (a) 0.41R
⛬ The height above the surface of the Earth where is weight is 30 kg is 0.41R