Question

Calculate the acceleration due to gravity at a point



a.64 km above and b)32 km below the surface of the earth
given radius of the earth= 6400km. Acceleration due to gravity on the surface of the earth g=9.8 ms^-2

Answer 1

Answer:

When an object moves in the gravitational field of the earth, it experiences a force called the gravitational force and the acceleration of the object is the acceleration due to gravity. Acceleration due to gravity near the surface of the earth has a constant values for every object i.e

g

=

9.8

m

/

s

2

.

The value of acceleration due to gravity changes if we go higher above the surface of the earth or go deep inside the surface of the earth. In either case the value of acceleration due to gravity decreases from its value on the surface of the earth.

Answer 2

(a) at height 64 km acceleration due to gravity becomes $9.6m/s^2$

(b) the acceleration due to gravity at depth 32 is $9.75m/s^2$.

Explanation:

(a)The acceleration due to gravity at height h is given as

$g`=\frac{g}{(1+\frac{h}{R})^2 }$

here g is acceleration due to gravity at earth surface and R is the radius of earth.

Given h = 64 km and R = 6400 km.

Substitute the values, we get

$g`=\frac{9.8m/s^2}{(1+\frac{64km}{6400km})^2 }$

$g`=9.6m/s^2$

Thus, at height 64 km acceleration due to gravity becomes $9.6m/s^2$

(b)the acceleration due to gravity at depth d is given by

$g`=g\frac{R-d}{R}$

Given d = 32 km,

so, $g`=9.8m/s^2(\frac{6400 km-32 km)}{6400km}$

$g`=9.75m/s^2$

Thus, the acceleration due to gravity at depth 32 is $9.75m/s^2$.

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