Question

The distance of the centres of moon and earth is D. The mass of earth is 81 times the mass of the moon. At what distance from the centres of the earth, the gravitational force will be zero?

Answer 1

Answer:

Given:

Distance between earth and moon = D

Mass of earth = 81m

Mass of moon = m

To find:

Distance from earth at which the gravitational force is zero.

Concept:

We will consider placing a mass M in between earth and moon . It will be placed at a distance "x" from the Earth.

Diagram:

Please refer to the attached photo to understand better.

Calculation:

F1 = F2

=> (GMm)/(D-x)² = GM(81m)/x²

=> 1/(D-x)² = 81/x²

=>1/(D-x) = 9/x. [square root on both sides]

=> x = 9D - 9x

=> 10x = 9D

=> x = (9/10)D.

At this distance from the centre os earth, the gravitational force will be zero.

Answer 2

${ \huge \bf { \mid{ \overline{ \underline{Question}}} \mid}}$

The distance of the centres of moon and earth is D. The mass of earth is 81 times the mass of the moon. At what distance from the centres of the earth, the gravitational force will be zero?

$\huge{\bold{\underline{\underline{Answer}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Let the Mass of moon be "m".
• Let the mass of Earth be "81 m" ( As the question specifies)
• Distance between Earth & Moon is "D".
• Let the Gravitational Force be zero at a point "r" from the Earth.

$\huge{\bold{\underline{Explanation:-}}}$

$\rule{300}{1.5}$

For Gravitational Force to Be zero.

Gravitational Force of Earth = Gravitational Force of Moon.

$\large{\boxed{\tt F_{[Earth]} = F_{[Moon]}}}$

Substituting the Formula,

$\large{\tt \leadsto \dfrac{GMm_1}{r_1^2} = \dfrac{GMm_2}{r_2^2}}$

$\large{\tt \leadsto \dfrac{\cancel{G M}m_1}{r_1^2} = \dfrac{\cancel{G M}m_2}{r_2^2}}$

$\large{\tt \leadsto \dfrac{m_1}{r_1^2} = \dfrac{m_2}{r_2^2}}$

• m₁ = Mass of Earth = 81 m
• m₂ = Mass of Moon = m.

$\large{\tt \leadsto \dfrac{81 m}{r_1^2} = \dfrac{m}{r_2^2}}$

$\large{\tt \leadsto \dfrac{81 \cancel{m}}{r_1^2} = \dfrac{\cancel{m}}{r_2^2}}$

$\large{\tt \leadsto \dfrac{81 }{r_1^2} = \dfrac{1}{r_2^2}}$

• r₁ = Distance from Earth = "r".
• r₂ = Distance from Moon = "D - r"

Substituting,

$\large{\tt \leadsto \dfrac{81 }{(r)^2} = \dfrac{1}{(D - r)^2}}$

Rearranging,

$\large{\tt \leadsto \dfrac{(D - r)^2}{r^2} = \dfrac{1}{81}}$

$\large{\tt \leadsto \bigg[ \dfrac{D - r}{r} \bigg]^2 = \dfrac{1}{81}}$

$\large{\tt \leadsto \bigg[ \dfrac{D - r}{r} \bigg] = \sqrt{\dfrac{1}{81}}}$

$\large{\tt \leadsto \dfrac{D - r}{r} = \dfrac{1}{9}}$

Cross - Multiplying it,

$\large{\tt \leadsto 9(D - r) = r}$

$\large{\tt \leadsto 9D - 9r = r}$

$\large{\tt \leadsto 9D = r + 9r}$

$\large{\tt \leadsto 9D = 10r}$

$\huge{\boxed{\boxed{\tt r = \dfrac{9D}{10}}}}$

Note:-

• Here I have taken the value of distance from the Earth as "r" you may take any other Letters to represent it like:- x, y, p, z ..... etc..

$\rule{300}{1.5}$