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.​

Answer 1

Given :

• Mass of the body = 25 kg

• Mass of the Earth = 6 × 10²⁴ kg

• Radius of the Earth = 6.4 × 10⁶ m

• Universal Gravitational constant = 6.7 × 10-¹¹ Nm²kg-²

To find :

• The force of Gravitation between the body and the Earth.

• Acceleration produced on the body.

• Acceleration produced on the Earth.

Solution :

To find the force of attraction between the body and the Earth :

We know the Equation for Force of Gravitation between the two bodies , i.e,

$\bf{F = G\dfrac{m_{1}m_{2}}{r^{2}}}$

Where :

• F = Force of Gravitation
• m = Mass
• r = Radius/distance

Using the above equation and substituting the values in it, we get :

$:\implies \bf{F = G\dfrac{m_{1}m_{2}}{r^{2}}}$

$:\implies \bf{F = 6.67 \times 10^{-11} \times \dfrac{25 \times 6 \times 10^{24}}{(6.4 \times 10^{6})^{2}}}$

$:\implies \bf{F = 6.67 \times 10^{-11} \times \dfrac{150 \times 10^{24}}{6.4^{2} \times (10^{6})^{2}}}$

$:\implies \bf{F = 6.67 \times 10^{-11} \times \dfrac{150 \times 10^{24}}{40.96 \times 10^{12}}}$

$:\implies \bf{F = 6.67 \times \dfrac{150 \times 10^{24 - 11}}{40.96 \times 10^{12}}}$

$:\implies \bf{F = 6.67 \times \dfrac{150 \times 10^{13}}{40.96 \times 10^{12}}}$

$:\implies \bf{F = 6.67 \times \dfrac{150 \times 10^{13 - 12}}{40.96}}$

$:\implies \bf{F = 6.67 \times \dfrac{150 \times 10^{1}}{40.96}}$

$:\implies \bf{F = 6.67 \times \dfrac{150 \times 10}{40.96}}$

$:\implies \bf{F = \dfrac{10005}{40.96}}$

$:\implies \bf{F = 244.26}$

$\boxed{\therefore \bf{Force\:(F) = 244.26\:N}}$

Hence, the force of Gravitation between the body and the Earth is 244.26 N.

To find the acceleration produced on the body :

We know that ,

$\bf{Force = mass \times Acceleration}$

So , using the above formula and substituting the values in it, we get :

$:\implies \bf{F = ma}$

$:\implies \bf{244.26 = 25 \times a}$

$:\implies \bf{\dfrac{244.26}{25} = a}$

$:\implies \bf{9.8 = a}$

$\boxed{\therefore \bf{Acceleration\:(a) = 9.8\:ms^{-2}}}$

Hence, the acceleration produced on the body is 9.8 m/s².

To find the acceleration produced on the Earth :

We know that ,

$\bf{Force = mass \times Acceleration}$

So , using the above formula and substituting the values in it, we get :

$:\implies \bf{F = ma}$

$:\implies \bf{244.26 = 6 \times 10^{24} \times a}$

$:\implies \bf{\dfrac{244.26}{6 \times 10^{24}} = a}$

$:\implies \bf{\dfrac{40.71}{10^{24}} = a}$

$:\implies \bf{40.71 \times 10^{-24} = a}$

$\boxed{\therefore \bf{Acceleration\:(a) = 40.71 \times 10^{-24}\:ms^{-2}}}$

Hence, the acceleration produced on the Earth is 40.71 × 10-²⁴ m/s²

Answer 2

Answer:

Given : mass of the body =25 kg

mass of the Earth =6×10^24 kg

Radius of the Earth=6.4×10^6 m

Gravitational constant=6.7×10^-11Nm^2kg^-2

1) Mutual force of attraction between the body and the Earth

F=Gm1m2/r^2

F= 6.7×10^-11×25×6×10^24 Nm^2/(6.4×10^6) ^2 m^2

F= 6.7×25×6×10^13N/40.96×10^12

F= 167.5×6 ×10^13×10^-12N/40.96

F=10005×10N/40.96

F= 244.26 N

so, the mutual force of attraction between the body and the Earth is 244.26N

2) acceleration produce by the body

since,F = ma

so, a= F/m

a=244.26/25

a=9.7705m/s^2 or 9.8 m/s^2

.

3) acceleration produced in the Earth

a=F/m

a=244.26/6×10^24

a= 40.71×10^-24m/s^2