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Question: If a graph is plotted between \({T^2}\) and \({r^3}\) for a planet, then its slope will be (where \(...

If a graph is plotted between T2{T^2} and r3{r^3} for a planet, then its slope will be (where MS{M_S} is the mass of the sun).
A) 4π2MS\dfrac{{4{\pi ^2}}}{{{M_S}}}
B) GMS4π\dfrac{{G{M_S}}}{{4\pi }}
C) 4πGMS4\pi G{M_S}
D) GMSG{M_S}

Explanation

Solution

Analyze the question by trying to find a relation between T2{T^2} and r3{r^3} following the motion of planets around the sun in their orbits. One of Keppler’s laws justifies such a relation. Here, TT is the time period of revolution of the planet around the Sun and rr is the radius of its orbit, i.e., distance from the Sun.

Complete step by step solution:
The three laws of Kepler about planetary motion are:
The Law of Ellipses: The path of the planets about the sun is elliptical, with the centre of the sun being located at one focus.
The Law of Equal Areas: An imaginary line drawn from the centre of the sun to the centre of the planet will sweep out equal areas in equal intervals of time.
The Law of Harmonies: The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun.
Since,
T=2πωT = \dfrac{{2\pi }}{\omega } where, ω\omega is the angular velocity of the body in circular motion
v=ωrv = \omega r
ω=vr\Rightarrow \omega = \dfrac{v}{r}
where, vv is the linear velocity of the body and rr is the radius of revolution
The orbital velocity of a planet revolving around the sun is given as
v=GMSr\Rightarrow v = \sqrt {\dfrac{{G{M_S}}}{r}}
Now,
T=2πrv\Rightarrow T = \dfrac{{2\pi r}}{v}
We put v=GMSrv = \sqrt {\dfrac{{G{M_S}}}{r}} and get-
T=2πrGMSr\Rightarrow T = \dfrac{{2\pi r}}{{\sqrt {\dfrac{{G{M_S}}}{r}} }}
where,
Ms{M_s} is mass of Sun as given
GG is the gravitational constant i.e., 6.673×1011Nm2kg26.673 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}
Squaring both sides,
T2=4π2r3GMS\Rightarrow {T^2} = \dfrac{{4{\pi ^2}{r^3}}}{{G{M_S}}}
Therefore, the slope of the graph would be 4π2MS\dfrac{{4{\pi ^2}}}{{{M_S}}} and it would be a straight-line graph as T2r3{T^2} \propto {r^3}

The correct answer is [A], 4π2MS\dfrac{{4{\pi ^2}}}{{{M_S}}}.

Note: Orbital velocity is the velocity at which a body revolves around another body in space. The orbital velocity is directly proportional to the mass of the body for which it is being calculated and inversely proportional to the radius.