Question

The height of the point vertically above the earth surface at which acceleration due to the garvity becomes 1 % of that value at the surface

Answer 1

Answer:

To find:

Height at which gravity acceleration becomes 1% to the value at the surface.

Formulas used:

Let height be h, radius of Earth be r, then gravity at height h is given as:

$g" = \frac{g}{ {(1 + \frac{h}{r}) }^{2} }$

Calculation:

As per the question, g" = (1/100) g

$\frac{1}{100} g= \frac{g}{ {(1 + \frac{h}{r}) }^{2} }$

$= > \frac{1}{100} = \frac{1}{ ({1 + \frac{h}{r} )}^{2} }$

$= > 1 + \frac{h}{r} = \sqrt{100}$

$= > 1 + \frac{h}{r} = 10$

$= > \frac{h}{r} = 9$

$= > h = 9r$

As per data , radius of Earth is 6400 km.

So, height = 9 × 6400 = 57600 km

$\boxed{ \boxed{height = 57600 \: km}}$

Answer 2

Answer:

${\boxed{\bold\green{Height =57600\:km}}}$

Explanation:

Given:-

Gravity becomes 1% of the value at the surface.

Let the new gravity be g".

Formula used:-

${\boxed{\bold\purple{g" =\frac{g}{{(1+\frac{h}{r})}^{2}}}}}$

${\boxed{\bold\green{r = 6400\:km}}}$

$\rightarrow\sf \frac{1}{100}\:g = \frac{g}{{(1+\frac{h}{r})}^{2}}$

$\rightarrow\sf \frac{g}{100} =\frac{g}{{(1+ \frac{h}{r})}^{2}}$

$\rightarrow\sf \frac{ 1}{100} = \frac{1}{{(1+\frac{h}{r})}^{2}}$

$\rightarrow\sf 100 = {(1+\frac{h}{r})}^{2}$

$\rightarrow\sf 1 + \frac{h}{r} = \sqrt{100}$

$\rightarrow\sf 1 + \frac{h}{r} = 10$

$\rightarrow\sf \frac{h}{r} = 10 - 1$

$\rightarrow\sf \frac{h}{r} = 9$

$\rightarrow\sf h = 9r$

$\rightarrow\sf h =( 9\times 6400) \:km \: \: \: [\because r = 6400\:km ]\\\\\rightarrow{\boxed{\bold\purple{h = 57600\:km}}}$

$\therefore\bold{\underline{Height = 57600\:km}}$