Question

a body weighs 25kg on the surface of the earth. if the mass of earth is 6*10^29 kg, radius is 6.4*10^6 m, G=6.67*10^-11Nm^2 / kg.calculate
(a) force of attraction between the body and earth.
(b) the acceleration produced in tje body.
(c) the acceleration produced in the earth.

Answer 1

24.42 * 10^6 newton of attraction.. 

Answer 2

Answer:

The force of attraction between the body and the earth is 2.4 ×$10^{7}$ N.

The acceleration produced in the body is 9.8 m/$s^{2}$

The acceleration brought about on Earth is 40.71 ×$10^{-24}$ m/s.

Explanation:

Given:

The body's mass (m) = 25 Kg

Earth's Mass (M) = 6 × $10^{29}$ Kg

Earth's Radius (r) = 6.4 ×$10^{6}$ m

The gravitational constant's value (G) = 6.67 ×$10^{-11}$ N$m^{2}$/kg

(i) The force of attraction (F) =

$F = \frac{GMm}{r^{2} }$

Putting all the given parameters in the given formula, we have

$F = \frac{6.67*10^{-11} *6*10^{29} *25}{(6.4*10^{6}) ^{2} }$

F = 244.26 N

Therefore, The force of attraction between the body and the earth is 2.4 ×$10^{7}$ N.

(b) F = ma

So,  244.26 = 25 × a

$a = \frac{244.6 }{25} = 9.8$

So, the acceleration produced in the body is 9.8 m/$s^{2}$

(c) F = ma

So, 244.26 = 6 ×$10^{24}$  × a

This implies, a = 40.71 ×$10^{-24}$ m/s.

Therefore, The acceleration brought about on Earth is 40.71 ×$10^{-24}$ m/s.

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