Question

A rocket is fired from the earth towards the moon. At what distance from the moon is the gravitational force on the rocket zero? Mass of earth = 6 x 1024 kg and Mass of moon = 7.4 x 1022 kg and orbital radius of the moon= 3.8 x 108 m. Neglect the effect of the sun and the other planets.

Answer 1

Answer: The distance from the moon is $d = 4 \times10^{7}m$.

Explanation:

Given that,

Mass of the earth $M_{e} = 6\times10^{24}kg$

Mass of the moon $M_{m} = 7.4\times10^{22}kg$

Radius of the moon $r_{m} = 3.8\times 10^{8}m$

Let us considered mass of the rocket is m.

We know that,

The gravitational force of the earth is

$F = \dfrac{GM_{e}m}{x^{2}}$

Where, x is the distance between the rocket and earth

The gravitational force of the moon

$F = \dfrac{GM_{m}m}{(r-x)^{2}}$

For rocket

$\dfrac{GM_{e}m}{x^{2}} = \dfrac{GM_{m}m}{(r-x)^{2}}$

$\dfrac{M_{e}}{x^{2}} =\dfrac{M_{m}}{(r-x)^{2}}$

$(r-x)^{2} = \dfrac{M_{m}}{M_{e}}x^{2}$....(I)

Put the values in equation (I)

$(3.8\times10^{8}-x)^{2} = (\dfrac{7.4\times10^{22}}{6\times10^{24}})x^{2}$

$3.8\times10^{8}-x= 0.11x$

$x = \dfrac{3.8\times10^{8}}{1.11}$

$x = 3.42\times10^{8}m$

Now, the distance from the moon is

$d = r-x$

$d = 3.8\times10^{8}m-3.42\times10^{8}m$

$d = 4 \times10^{7}m$

Hence, the distance from the moon is $d = 4 \times10^{7}m$.

Answer 2

Answer:

20.2 × 10¹⁹ N

Explanation:

Given :

Mass of earth = 6 × 10²⁴ kg

Mass of moon = 7.4 × 10²² kg

Distance between them = 3.84 × 10⁵ km = 3.84  × 10⁸ m

We have value of G = 6.7 × 10⁻¹¹ N m² kg⁻²

We have to find force :

We have :

F = G m₁ m₂ / r²

F = ( 6.7 × 10⁻¹¹ ) ( 6 × 10²⁴ ) ( 7.4 × 10²² ) / (  3.84 × 10⁸ )² N

F = 20.2 × 10¹⁹ N

Hence force exerted by the earth on the moon is 20.2 × 10¹⁹ N.