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Question: One of the satellites of Jupiter, has an orbital period of \[1.769\] days and the radius of the orbi...

One of the satellites of Jupiter, has an orbital period of 1.7691.769 days and the radius of the orbit is 4.22×108m4.22 \times {10^8}m . Show that the mass of Jupiter is about one-thousand that of the sun.

Explanation

Solution

Firstly we will find the value of mass of sun and Jupiter.After substituting the values of radius and time period of the Jupiter and sun. Then take the ratio of both the masses of Jupiter and sun.

Complete step by step solution:
Take Gravitational force equal to centripetal acceleration, we get
GmMr2=mv2r\dfrac{{GmM}}{{{r^2}}} = \dfrac{{m{v^2}}}{r}
v=GMrv = \sqrt {\dfrac{{GM}}{r}}
As we know,
T=2πωT = \dfrac{{2\pi }}{\omega }
Here T is the time period.
Also, v=ωrv = \omega r
So, ω=vr\omega = \dfrac{v}{r}
Time period, T=2πrv\dfrac{{2\pi r}}{v}
Squaring on both sides, we get,
T2=4π2r2v2{T^2} = \dfrac{{4{\pi ^2}{r^2}}}{{{v^2}}}
Now substitute the value of v in the resultant equation.
T2=4π2r3GM{T^2} = \dfrac{{4{\pi ^2}{r^3}}}{{GM}}
Here r is the radius, T is the time period, G is the gravitational force constant and M is the mass of the sun.

M=4π2r3GT2M = \dfrac{{4{\pi ^2}{r^3}}}{{G{T^2}}}
We know, radius of sun(1A.U=1.496×10111.496 \times {10^{11}}m), r=1.496×10111.496 \times {10^{11}}m and Time period is=365.25days365.25\,days.
Now substitute all the values, we get-
M=4π2(1.496×1011)3G(365.25×24×60×60)2M = \dfrac{{4{\pi ^2}{{\left( {1.496 \times {{10}^{11}}} \right)}^3}}}{{G{{\left( {365.25 \times 24 \times 60 \times 60} \right)}^2}}} ------ (1)

Now, the time period of Jupiter=1.7691.769 days=1.769×24×60×601.769 \times 24 \times 60 \times 60 s.
Radius of Jupiter=4.22×108m4.22 \times {10^8}m
So, mass of Jupiter will be MJ=4π2(4.22×108)3G(1.769×24×60×60)2{M_J} = \dfrac{{4{\pi ^2}{{\left( {4.22 \times {{10}^8}} \right)}^3}}}{{G{{\left( {1.769 \times 24 \times 60 \times 60} \right)}^2}}} ------ (2)
Now, divide equation (1) by (2),
MMJ=4π2(1.496×1011)3G(365.25×24×60×60)2×G(1.769×24×60×60)24π2(4.22×108)3\dfrac{M}{{{M_J}}} = \dfrac{{4{\pi ^2}{{\left( {1.496 \times {{10}^{11}}} \right)}^3}}}{{G{{\left( {365.25 \times 24 \times 60 \times 60} \right)}^2}}} \times \dfrac{{G{{\left( {1.769 \times 24 \times 60 \times 60} \right)}^2}}}{{4{\pi ^2}{{\left( {4.22 \times {{10}^8}} \right)}^3}}}
MMJ=1046\Rightarrow\dfrac{M}{{{M_J}}} = 1046
Here MM is the mass of the Sun and MJ{M_J} is the mass of Jupiter. We can write this equation as-
M1046=MJ\dfrac{M}{{1046}} = {M_J}
MJ11000M\therefore{M_J} \approx \dfrac{1}{{1000}}M

Hence proved, the mass of Jupiter is one-thousand times the mass of the sun.

Note: Jupiter is the fifth planet from the Sun and the largest in the solar system. It is a gas giant with a mass one-thousandth that of the sun. These two bodies affect one another proportionally according to distance and mass. Also Jupiter's gravity pulls on the sun is one-thousandth the amount the Sun’s gravity pulls on Jupiter.