CameraIcon

SearchIcon

MyQuestionIcon

2

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Acceleration Due to Gravity

Find the acce...

Question

Find the acceleration due to gravity in a mine of depth 640 m if the value at the surface is 9.800 ms^{-2} . The radius of the earth is 6400 km.

Open in App

Solution

Let g be the acceleration due to gravity in mine.

Then, $g^{'} = g (1 - \frac{d}{R}) = 9.8 (1 - \frac{640}{640 \times 10^{3}}) = 9.8 (\frac{10000 - 1}{10^{4}}) = \frac{9.8}{10^{4}} 9999 = 9.8 \times 0.9999 = 9.799 m / s^{2}$

Suggest Corrections

thumbs-up

12

Similar questions

Q. The value of acceleration due to gravity at a depth of

1600 k m

is equal to [Radius of earth

= 6400 k m

]

Q. The ratio of the acceleration due to gravity as the bottom of a deep mine and that on the surface of the earth is 978/980. Find the depth of the mine, if the density of the earth is uniform throughout and the radius of the earth is 6300 km.

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

g

2

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

6400 k m

). The place is:-

Q. The height at which the value of acceleration due to gravity becomes

50 %

of that at the surface of the earth

.

(radius of the earth =

6400 k m

g

on the surface of earth is

9.8 m / s^{2}

. Find acceleration due to gravity at depth

h = \frac{R}{2}

from the surface (

R =

radius of earth).

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Acceleration Due to Gravity

PHYSICS

Watch in App

explore_more_icon

Explore more

Acceleration Due to Gravity

Standard IX Physics

Join BYJU'S Learning Program

CrossIcon