a body weighs 25 kg on the surface of the earth. if mass od the earth is 6.0*10^{24} kg, the radius of the earth is 6.4*10^{6} m and the universal gravitational constant is 6.7*10^{-11}Nm^{2}/kg^{2}, calculate

- the force acting upon the body and the earth
- the acceleration produced in the body
- the acceleration produced in the earth

please help fast!!

It should be remebered that the gravitational force beween two bodies (A exerting on B and B on A) is always equal in magnitude but due to difference in the mass of A and B the magnitude of the effect of the force might differ, as in the case of an object kept on the surface of the earth.

(1)

The force acting upon the body of mass 'm' kept on earth'ssurface of mass 'M' will be

F = mg

where g is the acceleration due to gravity

so,

F = 25 x 9.8 = 245 N

and the force exerted by the body on earth will be

F' = (GMm) / r^{2}

where G is the gravitational constant and r is the radius of the earth

so,

F = (6.67x10^{-11} x 25 x 6x10^{24}) / (6.4x106)^{2}

or

F = 1.005 x 10^{16} / 40.96x10^{12}

thus,

F' = 244.26 ~ 245 N

thus, F = F'.

(2)

The acceleration produced in the body due to the gravitational pull of the earth will be equal to the acceleration due to gravity, g. So, it will be equal to 9.8 m/s^{2}.

(3)

The acceleration produced in the earth can be calculated by using this simple formula

a = F/M

here F will be the mutual gravitaional force and M is the mass of the earth.

so,

a = 245 / 6 x 1024

or

a = 4.083 x 10^{-23} m/s^{2}

which is negligibly small and for all purposes can be ignored.

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