Answer
Verified
91.5k+ views
Hint: The gravitation law states that every point mass attracts every other point mass by a force acting along the line intersecting the two points. The force is proportional to the product of the two masses, and inversely proportional to the square of the distance between them.
In equilibrium, the centrifugal force and the gravitational force will be equal.
Complete step by step solution:
Newton's law of universal gravitation is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
The force of gravity is a power that draws two mass objects. We call the force of gravity attractive because it constantly tries to draw masses together, never pushes them apart.
Let a planet of mass $m$, revolving in a circular orbit of radius $r$ around a massive star.
For equilibrium,
${F_ {centrifugal}} = {F_ {gravitation}} $
Where $F$ is force
According to the question,
${F_ {gravitation}}$ = $\dfrac{{GMm}}{r^{\dfrac{5}{2}}}$
Where, $G$ is gravitational constant, $M,m$ are the masses and $r$ is the radius.
Now, we know that
${F_ {gravitation}} = \dfrac{{m{v^2}}}{r}$
Where $v$ is the velocity
From the above two equations,
$\dfrac {{m {v^2}}} {r}$ = $\dfrac{{GMm}}{r^{\dfrac{5}{2}}}$
Solving,
${v^2} = \dfrac{{GM}}{{{r^{3.5}}}}$
Now, Time period is defined as
$T = \dfrac{{2\pi r}} {v} $
Therefore,
${T^2} = \dfrac{{4{\pi ^2} {r^2}}} {{{v^2}}} $
We know that
${v^2} = \dfrac{{GM}}{{{r^{3.5}}}}$
Thus,
${T^2} = \dfrac{{\dfrac{{4{\pi ^2} {r^2}}} {{GM}}}} {{{r^ {3.5}}}} $
Hence, ${T^2} $ is proportional to ${r^ {-3.5}} $.
The square of the time period will be proportional to ${r^ {-3.5}}$.
Note: The relation of the distance of objects in free fall to the square of the time taken had recently been confirmed by Grimaldi and Riccioli between 1640 and 1650. They had also made a calculation of the gravitational constant by recording the oscillations of a pendulum.
In equilibrium, the centrifugal force and the gravitational force will be equal.
Complete step by step solution:
Newton's law of universal gravitation is usually stated as that every particle attracts every other particle in the universe with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
The force of gravity is a power that draws two mass objects. We call the force of gravity attractive because it constantly tries to draw masses together, never pushes them apart.
Let a planet of mass $m$, revolving in a circular orbit of radius $r$ around a massive star.
For equilibrium,
${F_ {centrifugal}} = {F_ {gravitation}} $
Where $F$ is force
According to the question,
${F_ {gravitation}}$ = $\dfrac{{GMm}}{r^{\dfrac{5}{2}}}$
Where, $G$ is gravitational constant, $M,m$ are the masses and $r$ is the radius.
Now, we know that
${F_ {gravitation}} = \dfrac{{m{v^2}}}{r}$
Where $v$ is the velocity
From the above two equations,
$\dfrac {{m {v^2}}} {r}$ = $\dfrac{{GMm}}{r^{\dfrac{5}{2}}}$
Solving,
${v^2} = \dfrac{{GM}}{{{r^{3.5}}}}$
Now, Time period is defined as
$T = \dfrac{{2\pi r}} {v} $
Therefore,
${T^2} = \dfrac{{4{\pi ^2} {r^2}}} {{{v^2}}} $
We know that
${v^2} = \dfrac{{GM}}{{{r^{3.5}}}}$
Thus,
${T^2} = \dfrac{{\dfrac{{4{\pi ^2} {r^2}}} {{GM}}}} {{{r^ {3.5}}}} $
Hence, ${T^2} $ is proportional to ${r^ {-3.5}} $.
The square of the time period will be proportional to ${r^ {-3.5}}$.
Note: The relation of the distance of objects in free fall to the square of the time taken had recently been confirmed by Grimaldi and Riccioli between 1640 and 1650. They had also made a calculation of the gravitational constant by recording the oscillations of a pendulum.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
The vapour pressure of pure A is 10 torr and at the class 12 chemistry JEE_Main
Electric field due to uniformly charged sphere class 12 physics JEE_Main
3 mole of gas X and 2 moles of gas Y enters from the class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
Velocity of car at t 0 is u moves with a constant acceleration class 11 physics JEE_Main