Q. 44.0( 28 Votes )

# Io, one of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 10^{8} m. Show that the mass of Jupiter is about one-thousandth that of the sun.

Answer :

If a body orbits around other heavier body due to gravitational force of attraction then we have a relation between mass of heavier body, time period of revolution and, radius of the orbit as

Where M is the mass of heavier body, R is the radius of orbit, G is universal gravitational Constant and T is time period of revolution

Here we will consider two cases, motion of Satellite Io around Jupiter and motion of earth around sun

So for motion of Io around Jupiter we have

Where, M_{j} is the Mass of Jupiter

G is Universal gravitational constant

R_{Io} is radius of Io’s orbit, we are given

R_{Io} = 4.22 × 10^{8} m

T_{Io} is the time period of revolution of Io around Jupiter

TIo = 1.769 days

So for motion of Earth around Sun we have

Where, Ms is the Mass of sun

G is Universal gravitational constant

R_{e} is radius of Earth’s orbit

R_{e} = 1 AU = 1.496 × 10^{11} m

T_{e} is the time period of revolution of Earth around sun

T_{e} = 365.25 days

Diving equation of mass of sun with equation of mass of Jupiter to compare

So putting values of R_{e}, R_{Io}, T_{Io}, T_{e} in above equation we get

**NOTE:** Time periods are not converted to SI units second, but are in years, but would not make any difference to result because we are taking ration and units and converting factors are going to be cancelled out ultimately

Solving above equation we get

Or we can say

M_{s} ≈ 1000M_{j}

i.e. mass of Sun is nearly 1000 times mass of Jupiter

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A Saturn year is 29.5 times the earth year. How far is the Saturn from the sun if the earth is 1.50 × 10^{8} km away from the sun?

(a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth’s rotation.

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