Question

29. If mass of a planet is 2% of mass of earth. The ratio of
gravitational pull of earth on the planet and that of
planet on the earth will be
(a) 1 : 20 (b) 2:5 (c) 1:1 (d) 1 : 50
​

Answer 1

Answer:

Correct option is

A

1 : 1

Mass of earth be M and mass of moon be m.

Let the distance between moon and earth be r

Now force exerted by earth on moon, F

e

=

r

2

GMm

Force exerted by moon on earth F

m

=

r

2

GmM

F

e

:F

m

=1:1

This can also be solved by newton's 3rd law of motion.

Answer 2

C) 1:1 is the ratio of gravitational pull of the earth on the planet and gravitational pull of planet on the earth

Step-by-step Explanation:

Given: mass of the planet ($m_p$) = 2% mass of the earth ($m_e$)

To Find: ratio of gravitational pull of the earth on the planet and that of planet on the earth

Solution:

• The ratio of the gravitational pull of the earth on the planet and gravitational pull of planet on the earth

The gravitational force between the two bodies is mathematically expressed as:-

$F_g = G \frac{m_1 m_2}{r^{2}}$

Where $m_1$ and $m_2$ are the masses of the two bodies respectively, $r$ is the distance between them, and G is the gravitational constant.

Therefore, the gravitational pull of the earth on the planet is;

$(F_g)_{earth \rightarrow planet} = G \frac{m_e m_p}{r^{2}} \ \cdots \cdots (1)$

While the gravitational pull of the planet on the earth is;

$(F_g)_{planet \rightarrow earth} = G \frac{m_e m_p}{r^{2}} \ \cdots \cdots (2)$

Divide (1) by (2), you will get;

${\frac{(F_g)_{planet \rightarrow earth}}{(F_g)_{earth \rightarrow planet}} = \frac{G \frac{m_e m_p}{r^{2}}}{G \frac{m_e m_p}{r^{2}}}} = 1:1$

Hence, the ratio of the gravitational pull of the earth on the planet and that of the planet on the earth is 1:1