Question

A rocket moving around the earth at height “H”, If the gravitational acceleration “g1“ at height
“H” is ½ of gravitational acceleration “g“ at earth surface. If Earth radius is ”R”, find “H”
using R.​

Answer 1

ANSWER:

• The value of H = (√2 - 1) R.

GIVEN:

• A rocket moving around the earth at height H.

• The gravitational acceleration g' at height H is ½ of gravitational acceleration g at earth surface.

• Earth's radius is R.

TO FIND:

• The value of H in terms of R.

EXPLANATION:

$\bigstar\boxed{\bold{ \large{\red{g = \dfrac{GM}{R^2}}}}}\\ \\ \\ \bigstar\boxed{\bold{ \large{\pink{g' = \dfrac{GM}{(R+H)^2}}}}} \\ \\ \\ \sf \orange{ g' = \dfrac{g}{2}}\\ \\ \\ \sf \dfrac{\dfrac{GM}{R^2}}{2} = \dfrac{GM}{(R+H)^2}\\ \\ \\ \sf \dfrac{1}{R^2}= 2\left(\dfrac{1}{(R+H)^2}\right)\\ \\ \\ \textsf{\blue{Taking square root on both sides.}}\\ \\ \\ \sf \dfrac{1}{R}= \sqrt{2} \left(\dfrac{1}{R+H} \right)\\ \\ \\ \sf R + H = \sqrt{2}R\\ \\ \\ \sf H = (\sqrt{2} - 1)R$

Hence the value of H = (√2 - 1) R.

Answer 2

Given :-

A rocket moving around the earth at height “H”, If the gravitational acceleration “g1“ at height

“H” is ½ of gravitational acceleration “g“ at earth surface. If Earth radius is ”R”,

To Find :-

Value of H

Solution :-

We know that

$\sf g = \dfrac{GM}{R^2}$

Where

M = mass

G = acceleration

R = Radius

$\sf g` = \dfrac{GM}{(R+H)^2}$

Now

$\sf g` = \dfrac{g}{2}$

So,

$\sf g` = g`$

$\sf \dfrac{GM}{R^2} \div 2 = \dfrac{GM}{(R+H)^2}$

$\sf \dfrac{GM}{R^2} \times \dfrac{1}2 = \dfrac{GM}{(R+H)^2}$

$\sf \dfrac{1}{R^2} = 2 \times \dfrac{1}{(R+H)^2}$

$\sf\dfrac{1}{R^2} = \sqrt{2}\times \dfrac{1}{R+H}$

$\sf R\times\sqrt{2} = R+H$

$\sf\sqrt{2}R = R+H$

$\sf\sqrt{2}R - R=H$

$\sf R( \sqrt{2}-1) = H$