How far away from the centre of the earth does the acceleration due to gravity will be reduced by $1$ % of its value at the earth's surface?
Radius of earth $R = 6400 k m$ .

Open in App

Solution

at distance $r$ form center of earth the
value of $9$ will be frst $0.99 g$
we have
$g^{1} = \frac{g R^{2}}{r^{2}}$
given that $g^{1} = 0.99 g$
$\Rightarrow 0.99 g = g \frac{R^{2}}{r^{2}}$
$\Rightarrow \sqrt{0.99} = \frac{R}{r}$
$\Rightarrow r = \frac{R}{\sqrt{0.99}} = 1.005 R$
put $R = 6400 k m$
we get $r = 6432 k m$

Suggest Corrections

Similar questions

Q. How far away from the surface of the earth does the acceleration due to gravity becomes

4 %

of its value on the surface of earth?

[R_{e} = 6400 k m]

Hint:

g_{1} = g [\frac{R}{R + h}]^{2}

Q. How far away from the earth does the acceleration due to gravity become

10 %

of its value on earth's surface? Radius of earth

= 6.37 \times 10^{6}

Q. How much the surface of earth does the acceleration due to gravity reduce by

36 %

of its value on the surface of earth? Radius of earth

= 6400

km.

Q. At a place, value of acceleration due to gravity

g

is reduced by

2

% of its value on the surface of the earth (Radius of earth =

6400 k m

). The place is:-

The height above the Earth's surface at which the value of acceleration due to gravity reduces to half its value on the Earth's surface is h. Assume the Earth to be a sphere of radius 6400 Km. Find h.

Join BYJU'S Learning Program

How far away from the centre of the earth does the acceleration due to gravity will be reduced by 1% of its value at the earth's surface?Radius of earth R=6400km.

at distance r form center of earth the value of 9 will be frst 0.99 gwe have g1=gR2r2given that g1=0.99 g⇒0.99 g=gR2r2⇒√0.99=Rr⇒r=R√0.99=1.005 Rput R=6400 kmwe get r=6432 km

How far away from the centre of the earth does the acceleration due to gravity will be reduced by $1$ % of its value at the earth's surface?
Radius of earth $R = 6400 k m$ .