Question

A body weighs 25kg on the surface of the Earth. if the mass of the earth is 6×10²⁴ kg ,the radius of

the earth is 6.4×10⁶m and the gravitational constant 6.7×10-¹¹Nm²kg-².

Calculate-

1) Mutual force of attraction between the body and the earth.

2) The acceleration produced in the body.

3) The acceleration produced in the earth.
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Answer 1

Given:

A body weighs 25kg on the surface of the Earth. if the mass of the earth is 6×10²⁴ kg ,the radius of

the earth is 6.4×10⁶m and the gravitational constant 6.7×10-¹¹Nm²kg-².

To find:

1) Mutual force of attraction between the body and the earth.

2) The acceleration produced in the body.

3) The acceleration produced in the earth.

Calculation:

Applying Newton's Law of Gravitation, let mutual force be F ;

$\therefore \: F = \dfrac{G(m1)(m2)}{ {r}^{2} }$

$= > \: F = \dfrac{(6.7 \times {10}^{ - 11})(25)(6 \times {10}^{24} )}{ {(6.4 \times {10}^{6} )}^{2} }$

$= > \: F = \dfrac{(6.7 \times {10}^{ - 11})(25)(6 \times {10}^{24} )}{ 40.96 \times {10}^{12} }$

$= > \: F = \dfrac{(6.7 \times {10}^{ - 11})(25)(6 \times {10}^{12} )}{ 40.96 }$

$= > \: F = \dfrac{(6.7 \times 10)(25)(6 )}{ 40.96 }$

$= > \: F = \dfrac{(67)(25)(6 )}{ 40.96 }$

$\boxed{= > \: F = 245.36 \: newton}$

Now , acceleration in the body :

$\therefore \: a1 = \dfrac{F}{m1}$

$= > \: a1 = \dfrac{245.36}{25}$

$\boxed{= > \: a1 = 9.814 \: m {s}^{ - 2} }$

Acceleration produced in earth :

$\therefore \: a2= \dfrac{F}{m2}$

$= > \: a2 = \dfrac{245.36}{6 \times {10}^{24} }$

$\boxed{ = > \: a2 = 40.89 \times {10}^{ - 24} \: m {s}^{ - 1} }$

HOPE IT HELPS.