
If the distance between centres of the earth and the moon is $d$ and the mass of the earth is $81$ times the mass of the moon, then at what distance from centre of the earth, the gravitational field will be zero?
(A) $\dfrac{d}{2}$
(B) $\dfrac{d}{9}$
(C) $\dfrac{{9d}}{{10}}$
(D) $\dfrac{d}{{10}}$
Answer
535.2k+ views
Hint: In the given question, consider the particle to be placed at a distance from the earth. Now the gravitational force on the particle will be zero when the force on the particle due to earth equals the force on the particle due to the moon. Using this relation, we can get the required solution.
Complete Step-By-Step Solution:
It is given in the question, it is given that the distance between the moon and the earth is$d$.
Let us consider a particle of mass$m$, is placed at a distance $x$ from the earth.
Therefore, the distance of the particle from the moon is $d - x$
Therefore, we know that force exerted by one body due to another is given by:
$F = \dfrac{{G{M_1}{M_2}}}{{{r^2}}}$
Where,
$F$is the gravitational force,
$G$ is the gravitational constant
${M_1}$ and ${M_2}$be the mass of the bodied respectively
$r$ be the distance between the bodies.
Now, let us consider mass of the moon be $M$, then according to the question, mass of the earth is $81M$
Thus, we can write
Force exerted by the earth on the particle is given by:
${F_E} = \dfrac{{G81Mm}}{{{x^2}}}$
And, force exerted by moon the particle is given by:
${F_M} = \dfrac{{GMm}}{{{{(d - x)}^2}}}$
We know that gravitational force on the particle will be zero when:
${F_M} = {F_E}$
Thus, putting the values, we obtain:
$\dfrac{{GMm}}{{{{(d - x)}^2}}} = \dfrac{{G81Mm}}{{{{(x)}^2}}}$
On cancelling the common terms and solving the equation, we get:
$81{(d - x)^2} = {(x)^2}$
$ \Rightarrow x = \dfrac{{9d}}{{10}}$
This is our required solution.
Hence, option (C ) is correct.
Note:
According to Newton’s universal law of gravitation, the gravitational force between two bodies is directly proportional to the masses of the bodies and inversely proportional to the square of the distance between them. Thus, the constant of proportionality in the universal law of gravitation is known as the Gravitational constant.
Complete Step-By-Step Solution:
It is given in the question, it is given that the distance between the moon and the earth is$d$.
Let us consider a particle of mass$m$, is placed at a distance $x$ from the earth.
Therefore, the distance of the particle from the moon is $d - x$
Therefore, we know that force exerted by one body due to another is given by:
$F = \dfrac{{G{M_1}{M_2}}}{{{r^2}}}$
Where,
$F$is the gravitational force,
$G$ is the gravitational constant
${M_1}$ and ${M_2}$be the mass of the bodied respectively
$r$ be the distance between the bodies.
Now, let us consider mass of the moon be $M$, then according to the question, mass of the earth is $81M$
Thus, we can write
Force exerted by the earth on the particle is given by:
${F_E} = \dfrac{{G81Mm}}{{{x^2}}}$
And, force exerted by moon the particle is given by:
${F_M} = \dfrac{{GMm}}{{{{(d - x)}^2}}}$
We know that gravitational force on the particle will be zero when:
${F_M} = {F_E}$
Thus, putting the values, we obtain:
$\dfrac{{GMm}}{{{{(d - x)}^2}}} = \dfrac{{G81Mm}}{{{{(x)}^2}}}$
On cancelling the common terms and solving the equation, we get:
$81{(d - x)^2} = {(x)^2}$
$ \Rightarrow x = \dfrac{{9d}}{{10}}$
This is our required solution.
Hence, option (C ) is correct.
Note:
According to Newton’s universal law of gravitation, the gravitational force between two bodies is directly proportional to the masses of the bodies and inversely proportional to the square of the distance between them. Thus, the constant of proportionality in the universal law of gravitation is known as the Gravitational constant.
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