The distance of a planet from the sun is $$5$$ times the distance between the earth and the sun. The time period of the planet is:

 1 $$5^{3/2}$$ years 2 $$5^{2/3}$$ years 3 $$5^{1/3}$$ years 4 $$5^{1/2}$$ years

Subtopic:  Kepler's Laws |
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A satellite whose mass is $$m$$, is revolving in a circular orbit of radius $$r$$, around the earth of mass $$M$$. Time of revolution of the satellite is:
1. $$T \propto \frac{r^5}{GM}$$
2. $$T \propto \sqrt{\frac{r^3}{GM}}$$
3. $$T \propto \sqrt{\frac{r}{\frac{GM^2}{3}}}$$
4. $$T \propto \sqrt{\frac{r^3}{\frac{GM^2}{4}}}$$

Subtopic:  Kepler's Laws |
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A planet moves around the sun. At a point $$P$$, it is closest to the sun at a distance $$d_1$$ and has speed $$v_1.$$ At another point $$Q$$, when it is farthest from the sun at distance $$d_2,$$ its speed will be:

 1 $$d_2v_1 \over d_1$$ 2 $$d_1v_1 \over d_2$$ 3 $$d_1^2v_1 \over d_2$$ 4 $$d_2^2v_1 \over d_1$$
Subtopic:  Kepler's Laws |
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Kepler's third law states that the square of the period of revolution ($$T$$) of a planet around the sun, is proportional to the third power of average distance $$r$$ between the sun and planet i.e. $$T^2 = Kr^3$$, here $$K$$ is constant. If the masses of the sun and planet are $$M$$ and $$m$$ respectively, then as per Newton's law of gravitation, the force of attraction between them is $$F = \frac{GMm}{r^2},$$ here $$G$$ is the gravitational constant. The relation between $$G$$ and $$K$$ is described as:
1. $$GK = 4\pi^2$$
2. $$GMK = 4\pi^2$$
3. $$K =G$$
4. $$K = \frac{1}{G}$$

Subtopic:  Kepler's Laws |
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NEET - 2015
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If $$A$$ is the areal velocity of a planet of mass $$M,$$ then its angular momentum is:

 1 $$\frac{M}{A}$$ 2 $$2MA$$ 3 $$A^2M$$ 4 $$AM^2$$
Subtopic:  Kepler's Laws |
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If $$R$$ is the radius of the orbit of a planet and $$T$$ is the time period of the planet, then which of the following graphs correctly shows the motion of a planet revolving around the sun?

 1 2 3 4
Subtopic:  Kepler's Laws |
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A satellite that is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of the earth in the new orbit is $$2$$ times that of the earlier orbit. The time period in the second orbit is:
1. $$48$$$\sqrt{2}$ hr
2. $$48$$ hr
3. $$24$$$\sqrt{2}$ hr
4. $$24$$ hr

Subtopic:  Kepler's Laws |
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The kinetic energies of a planet in an elliptical orbit around the Sun, at positions $$A,B~\text{and}~C$$ are $$K_A, K_B~\text{and}~K_C$$ respectively. $$AC$$ is the major axis and $$SB$$ is perpendicular to $$AC$$ at the position of the Sun $$S$$, as shown in the figure. Then:

 1 $$K_A K_B> K_C$$ 3 $$K_B K_A> K_C$$
Subtopic:  Kepler's Laws |
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NEET - 2018
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The figure shows the elliptical orbit of a planet $$m$$ about the sun $$\mathrm{S}.$$ The shaded area $$\mathrm{SCD}$$ is twice the shaded area $$\mathrm{SAB}.$$ If $$t_1$$ is the time for the planet to move from $$\mathrm{C}$$ to $$\mathrm{D}$$ and $$t_2$$ is the time to move from $$\mathrm{A}$$ to $$\mathrm{B},$$ then:

 1 $$t_1>t_2$$ 2 $$t_1=4t_2$$ 3 $$t_1=2t_2$$ 4 $$t_1=t_2$$

Subtopic:  Kepler's Laws |
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AIPMT - 2009
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If two planets are at mean distances $$d_1$$ and $$d_2$$ from the sun and their frequencies are $$n_1$$ and $$n_2$$ respectively, then:
1. $$n^2_1d^2_1= n_2d^2_2$$
2. $$n^2_2d^3_2= n^2_1d^3_1$$
3. $$n_1d^2_1= n_2d^2_2$$
4. $$n^2_1d_1= n^2_2d_2$$

Subtopic:  Kepler's Laws |
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