# NCERT Solution : Gravitation

Question : 1. Suppose there existed a planet that went around the sun twice the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth ?

Solution: From Kepler’s 3rd law ,

$\displaystyle \frac{T_1^2}{T_2^2} = \frac{R_1^3}{R_2^3}$

Here, T1 = 1 years, T2 = 1/2 year, R1 = 1A.U. , R2 = ?

$\displaystyle R_2 = R_1 (\frac{T_2}{T_1})^{2/3} = 1(\frac{1/2}{1})^{2/3}$

R2 = 0.63 A.U

Question :2. One of the satellites of Jupiter, has an orbital period of 1.769 days and the radius of the orbit is 4.22 × 108 m. show that mass of Jupiter is about one thousand times that of sun. (Take 1 year = 365.25 mean solar day)

Solution: For the satellite of Jupiter,

T1 = 1.769 days

= 1.769 × 24 × 60 × 60 s

R1 = 4.22 × 108 m

$\displaystyle \frac{G M m}{R^2} = m\omega^2 R$

$\displaystyle \frac{G M }{R^2} = \omega^2 R$

$\displaystyle \frac{G M }{R^2} = (\frac{2\pi}{T})^2 R$

$\displaystyle M = \frac{4 \pi^2 R^3}{T^2 G}$

Mass of Jupiter , $\displaystyle M_1 = \frac{4 \pi^2 R_1^3}{T_1^2 G}$

$\displaystyle M_1 = \frac{4\times (3.14)^2 (4.22\times 10^8)^3}{(1.769\times 24 \times 60\times 60)^2 \times 6.67 \times 10^{-11}}$ …(i)

For earth around the sun, T2 = 1 year = 365.25 × 24 × 60 × 60 s

R2 = 1 A.U = 1.496 × 1011 m

Mass of the sun , $\displaystyle M_2 = \frac{4 \pi^2 R_2^3}{T_2^2 G}$

$\displaystyle M_2 = \frac{4\times (3.14)^2 (1.496\times 10^11)^3}{(365.25\times 24 \times 60\times 60)^2 \times 6.67 \times 10^{-11}}$ …(ii)

Dividing (ii) by (i)

$\displaystyle \frac{M_2}{M_1} = 1046$

Question : 3.Let us consider that our galaxy consists of 2.5 × 1011 stars each of one solar mass. How long will this star at a distance of 50,000 ly from the galactic centre take to complete one revolution ? Take the diameter of the Milky way to be 105 ly. G = 6.67 × 10-11 Nm2/kg2

Solution: r = 50,000 ly

= 50,000 × 9.46 × 1015 ,

m = 4.73 × 1020 m

M = 2.5 × 1011 solar mass

= 2.5 × 1011 × 2 × 1030 kg

= 5 × 1041 kg

$\displaystyle M = \frac{4 \pi^2 r^3}{G T^2 }$

$\displaystyle T = (\frac{4 \pi^2 r^3}{ G M})^{1/2}$

$\displaystyle T = (\frac{4 (3.14)^2 (4.73 \times 10^{20})^3}{ (6.67\times 10^{-11})(5 \times 10^{41} })^{1/2}$

= 1.12 × 1016 s

Question : 4. A rocket is fired from the earth towards the sun. At what point on its path is the gravitational force on the rocket zero? Mass of sun = 2 × 103 kg, Mass of earth = 6 × 1024 kg. Neglect the effect of the other planets. Orbital radius of the earth = 1.5 × 1011 m

Solution: Ms = 2 × 1030 kg.

Me = 6 × 1024 kg , r = 1.5 × 1011 m

Let at a distance x from the earth the gravitational force on the rocket due to sun and the earth are equal and opposite.

Distance of rocket from the sun = r – x

If m is the mass of the rocket then

$\displaystyle \frac{G M_s m}{(r-x)^2} = \frac{G M_e m}{x^2}$

$\displaystyle \frac{M_s}{M_e} = (\frac{r-x}{x})^2$

$\displaystyle \frac{r-x}{x} = \sqrt{\frac{M_s}{M_e}}$

$\displaystyle \frac{r-x}{x} = \sqrt{\frac{2\times 10^{30}}{6\times 10^{24}}}$

x = 2.59 × 108 m

Question: 5.A Saturn year is 29.5 times the earth year. How far is the Saturn from the Sun if the Earth is 1.5 × 108 km away from the sun ?

Solution: Ts = 29.5 Te ; Re = 1.5 × 108 km;

From Kepler’s 3rd law

$\displaystyle \frac{T_s^2}{T_e^2} = \frac{R_s^3}{R_e^3}$

$\displaystyle R_s = R_e (\frac{T_s}{T_e})^{2/3}$

$\displaystyle R_s = 1.58\times 10^8(\frac{29.5 Te}{T_e})^{2/3}$

= 1.43 × 109 km

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