The value of g will become $10 %$ of its value at the earth's surface at a height _____ above the surface of Earth.

Open in App

Solution

Step 1: Given quantities:
$g^{'} = 10 % o f g g^{'} = 0.1 g g^{'} \to a c c e l e r a t i o n d u e t o g r a v i t y a t a h e i g h t g \to a c c e l e r a t i o n d u e t o g r a v i t y a t t h e s u r f a c e$
Step 2: Formula Used:
$\frac{g^{'}}{g} = {(\frac{R_{e}}{R_{e} + h})}^{2} R_{e} \to r a d i u s o f E a r t h h \to h e i g h t f r o m t h e s u r f a c e$
Step 3: The calculation for the height:
Putting the values in the formula, we get
$\frac{0.1 g}{g} = {(\frac{6400}{6400 + h})}^{2} 0.1 = {(\frac{6400}{6400 + h})}^{2} \frac{6400}{6400 + h} = \sqrt{0.1} \frac{6400}{\sqrt{0.1}} = 6400 + h h = \frac{6400}{\sqrt{0.1}} - 6400 h = 13838 K m$
Hence, The value of g will become $10 %$ of its value at the earth's surface at a height $13838 K m$ above the surface of Earth.

Suggest Corrections

Similar questions

The value of g will become 1% of its value at the earth's surface at the height ________ above the surface of the earth. Take the radius of the earth as 6400km

Q. At what height above the earth's surface does the value of g becomes 36% of the value at the surface of earth ?

Q. The value of

g

on the earths surface is

980 c m / {s e c}^{2}

. Its value at a height of

64 K m

from the earth's surface is

Q. The height of the point vertically above the earth's surface at which acceleration due to gravity becomes

1

% of its value at the surface of earth is

(R =

radius of earth

)

Q. The height vertically above the earth's surface at which the acceleration duo to gravity becomes 1% of its value at the surface is (It is the radius of the earth)

Join BYJU'S Learning Program

The value of g will become 10% of its value at the earth's surface at a height _____ above the surface of Earth.

The value of g will become $10 %$ of its value at the earth's surface at a height _____ above the surface of Earth.