Question

18. The distance of the moon from the earth is
3.8 x 105 km. Calculate the speed of the moon
revolving around the earth. Mass of the earth
6.1 x 1024 kg and G = 6.7 x 10-11 N m² kg ?

Answer 1

Correct Question:-

The distance of the moon from the earth is 3.8 x 10^5 km. Calculate the speed of the moon revolving around the earth. Mass of the earth is 6.1 x 10^24 kg and G = 6.7 x 10^ -11 Nm²/kg².

Given data:-

The distance of the moon from the earth is 3.8 x 10^5 km.

Mass of the earth 6.1 x 10^24 kg and G = 6.7 x 10^ -11 Nm²/kg²

Solution:-

Here,

—› Mass of moon = m

—› Mass of earth = M

—› Gravitational constant = G (6.7 × 10^-11 Nm²/kg²)

—› Distance from moon to earth = r

—› Velocity of moon = v

—› Gravitational force = F

Now,

The moon is rotating around around the earth so in the moon their is a centrifugal force acting toward earth is given as

${ \dashrightarrow{ \bf{F \: = m \frac{ {v}^{2} }{r} }} \: \: \: \: \: \: \: .....(1)}$

Now, we know, according to Newton law of gravitation. earth exert a gravitatinal force ... is given by

${ \dashrightarrow{ \bf{ F= G \frac{Mm}{ {r}^{2} } }} \: \: \: \: \: .....(2)}$

Now, from eq. ( 1 ) & eq. ( 2 )

${ \dashrightarrow{ \bf{ m \frac{ {v}^{2} }{r} = G \frac{Mm}{ {r}^{2} } }} }$

So now,

${ \dashrightarrow{ \bf{ {v}^{2} = \frac{GM}{ {r}} }} }$

${ \dashrightarrow{ \bf{ {v} = \sqrt{\frac{GM}{ {r}} } }} }$

Now, from given

${ \dashrightarrow{ \bf{ {v} = \sqrt{\frac{6.7 \times {10}^{ - 11} \times \: 6.1 \times {10}^{24} }{3.8 \times {10}^{5} } } }} }$

${ \dashrightarrow{ \bf{ {v} = \sqrt{\frac{4.087 \times {10}^{14} }{3.8 \times {10}^{5} } } }} }$

${ \dashrightarrow{ \bf{ {v} = 32795.21788 \: \: m/s }} }$

Hence, the speed of the moon revolving around the earth is 32795.21788 m/s².

{Note: we need to take constant value for perfect velocity of moon = 1.022 km/s where, constant values are M = 5.972 × 10^24 kg ; G = 6.673 × 10^ - 11 Nm²/kg²; r = 3.84 × 10^8 m}