Given,

Orbital period of Io, T_{I0} = 1.769 days

= 1.769 × 24 × 60 × 60 s

Orbital radius of Io, R_{I0} = 4.22 × 10^{8} m

We know the mass of Jupiter:

M_{J} = 4π^{2}R_{I0}^{3} / GT_{I0}^{2}. . . . . . . . . . . . (1)

Where;

M_{J} = Mass of Jupiter

G = Universal gravitational constant

Also,

The orbital period of the earth,

T_{E }= 365.25 days = 365.25 × 24 × 60 × 60 s

Orbital radius of the Earth, R_{E }= 1 AU = 1.496 × 10^{11 }m

We know that the mass of sun is:

M_{S} =\( \dfrac{4\;\pi ^{2}\;R_{E}^{3}}{G\;T_{E}^{2}}\). . . . . . . . . . . . . (2)

Therefore, \( \\\dfrac{M_{S}}{M_{J}} = \dfrac{4\;\pi ^{2}\;R_{E}^{3}}{G\;T_{E}^{2}}\;\times \;\dfrac{G\;T_{10}^{2}}{4\;\pi ^{2}\;R_{10}^{3}}\)

\( = \dfrac{R_{E}^{3}}{T_{E}^{2}}\;\times \;\dfrac{T_{10}^{2}}{R_{10}^{3}}\)

Now, on substituting the values, we will get:

= \( \left [ \dfrac{1.769\times 24\times 60\times 60}{365.25\times 24\times 60\times 60} \right ]^{2}\)

= \( \times \left [ \dfrac{1.496\times 10^{11}}{4.22\times 10^{8}} \right ]^{3}\)

Therefore, \( \dfrac{M_{S}}{M_{J}} ~ 1000\)

M_{S }~ 1000 × M_{J }

[Which proves that, the Sun’s mass is 1000 times that of Jupiter’s]

Answered by Abhisek | 1 year agoA rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s^{–1}. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×10^{23} kg; radius of mars = 3395 km; G = 6.67×10^{-11} N m^{2} kg^{–2}.

A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the sun = 2×1030 kg; mass of mars = 6.4×1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 ×108 km; G = 6.67×10-11 N m2kg–2.

A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2×10^{30} kg).

As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite? (Take the potential energy at infinity to be zero). Mass of the earth = 6.0×10^{24} kg, radius = 6400 km

Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?