Answer

Verified

450.9k+ views

Hint – In this question let ${m_1}$ be the mass of the first body and the mass of the second celestial body be${m_2}$. Use the direct relationship between the force, masses and the distance between two bodies that is ${F_g} = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$. Since the masses are now changed therefore application of the same formula will help yielding two different equations. Use them to get the answer.

Step by step answer:

Let the first celestial body have mass ${m_1}$ and the second celestial body having mass${m_2}$.

Let them are separated by the distance r as shown in the figure.

As we know that the gravitational force (${F_g}$) working between these two bodies is given as

$ \Rightarrow {F_g} = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$...................... (1)

Where, G = universal gravitational constant.

Now it is given that the mass of any one body is doubled and the mass of the other body is halved and the gravitational force working between them is one-fourth of the previous one.

Let mass of first body = ${m_3}$

And mass of other body = ${m_4}$

Let the force between them = ${F'_g}$

Therefore, ${m_3} = 2{m_1}$

And ${m_4} = \dfrac{1}{2}{m_2}$

And, ${F'_g} = \dfrac{1}{4}{F_g}$

Let the distance between them is (d).

So the gravitational force between them is

$ \Rightarrow {F'_g} = G\dfrac{{{m_3}{m_4}}}{{{d^2}}}$

Now substitute the values we have,

$ \Rightarrow \dfrac{{{F_g}}}{4} = G\dfrac{{2{m_1}\left( {\dfrac{1}{2}{m_2}} \right)}}{{{d^2}}}$

$ \Rightarrow \dfrac{{{F_g}}}{4} = G\dfrac{{{m_1}\left( {{m_2}} \right)}}{{{d^2}}}$............. (2)

Now divide equation (1) from equation (2) we have,

$ \Rightarrow \dfrac{{{F_g}}}{{\dfrac{{{F_g}}}{4}}} = \dfrac{{G\dfrac{{{m_1}{m_2}}}{{{r^2}}}}}{{G\dfrac{{{m_1}\left( {{m_2}} \right)}}{{{d^2}}}}}$

Now simplify this equation we have,

$ \Rightarrow 4 = \dfrac{{{d^2}}}{{{r^2}}}$

Now take square root on both sides we have,

$ \Rightarrow \sqrt 4 = \sqrt {\dfrac{{{d^2}}}{{{r^2}}}} $

$ \Rightarrow \dfrac{d}{r} = 2$

$ \Rightarrow d = 2r$

So the separation between the bodies should be doubled.

So this is the required answer.

Note – There is often a confusion between g and G. g is the acceleration due to gravity whose value is 9.8 at the surface of the earth however G is the proportionality constant and has a default value of $6.674 \times {10^{ - 11}}{m^3}K{g^{ - 1}}{s^{ - 2}}$. It is advised to remember the direct formula for the force of gravitation between two masses that is ${F'_g} = G\dfrac{{{m_3}{m_4}}}{{{d^2}}}$.

Step by step answer:

Let the first celestial body have mass ${m_1}$ and the second celestial body having mass${m_2}$.

Let them are separated by the distance r as shown in the figure.

As we know that the gravitational force (${F_g}$) working between these two bodies is given as

$ \Rightarrow {F_g} = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$...................... (1)

Where, G = universal gravitational constant.

Now it is given that the mass of any one body is doubled and the mass of the other body is halved and the gravitational force working between them is one-fourth of the previous one.

Let mass of first body = ${m_3}$

And mass of other body = ${m_4}$

Let the force between them = ${F'_g}$

Therefore, ${m_3} = 2{m_1}$

And ${m_4} = \dfrac{1}{2}{m_2}$

And, ${F'_g} = \dfrac{1}{4}{F_g}$

Let the distance between them is (d).

So the gravitational force between them is

$ \Rightarrow {F'_g} = G\dfrac{{{m_3}{m_4}}}{{{d^2}}}$

Now substitute the values we have,

$ \Rightarrow \dfrac{{{F_g}}}{4} = G\dfrac{{2{m_1}\left( {\dfrac{1}{2}{m_2}} \right)}}{{{d^2}}}$

$ \Rightarrow \dfrac{{{F_g}}}{4} = G\dfrac{{{m_1}\left( {{m_2}} \right)}}{{{d^2}}}$............. (2)

Now divide equation (1) from equation (2) we have,

$ \Rightarrow \dfrac{{{F_g}}}{{\dfrac{{{F_g}}}{4}}} = \dfrac{{G\dfrac{{{m_1}{m_2}}}{{{r^2}}}}}{{G\dfrac{{{m_1}\left( {{m_2}} \right)}}{{{d^2}}}}}$

Now simplify this equation we have,

$ \Rightarrow 4 = \dfrac{{{d^2}}}{{{r^2}}}$

Now take square root on both sides we have,

$ \Rightarrow \sqrt 4 = \sqrt {\dfrac{{{d^2}}}{{{r^2}}}} $

$ \Rightarrow \dfrac{d}{r} = 2$

$ \Rightarrow d = 2r$

So the separation between the bodies should be doubled.

So this is the required answer.

Note – There is often a confusion between g and G. g is the acceleration due to gravity whose value is 9.8 at the surface of the earth however G is the proportionality constant and has a default value of $6.674 \times {10^{ - 11}}{m^3}K{g^{ - 1}}{s^{ - 2}}$. It is advised to remember the direct formula for the force of gravitation between two masses that is ${F'_g} = G\dfrac{{{m_3}{m_4}}}{{{d^2}}}$.

Recently Updated Pages

Mark and label the given geoinformation on the outline class 11 social science CBSE

When people say No pun intended what does that mea class 8 english CBSE

Name the states which share their boundary with Indias class 9 social science CBSE

Give an account of the Northern Plains of India class 9 social science CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

Advantages and disadvantages of science

Trending doubts

Bimbisara was the founder of dynasty A Nanda B Haryanka class 6 social science CBSE

Which are the Top 10 Largest Countries of the World?

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

10 examples of evaporation in daily life with explanations

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

How do you graph the function fx 4x class 9 maths CBSE

Difference Between Plant Cell and Animal Cell