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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 |

80%

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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. $\mathrm{T}\propto \frac{{\mathrm{r}}^{5}}{\mathrm{GM}}$

2. $\mathrm{T}\propto \sqrt{\frac{{\mathrm{r}}^{3}}{\mathrm{GM}}}$

3. $\mathrm{T}\propto \sqrt{\frac{\mathrm{r}}{{\mathrm{GM}}^{2}/3}}$

4. $\mathrm{T}\propto \sqrt{\frac{{\mathrm{r}}^{3}}{{\mathrm{GM}}^{2}/4}}$

Subtopic: Kepler's Laws |

81%

From NCERT

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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 |

81%

From NCERT

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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 |

79%

From NCERT

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 |

74%

From NCERT

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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 |

78%

From NCERT

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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 |

74%

From NCERT

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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. | \(K_A <K_B< K_C\) | 2. | \(K_A >K_B> K_C\) |

3. | \(K_B <K_A< K_C\) | 4. | \(K_B >K_A> K_C\) |

Subtopic: Kepler's Laws |

76%

From NCERT

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 |

71%

From NCERT

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. ${{n}_{1}}^{2}{{d}_{1}}^{2}={n}_{2}{{d}_{2}}^{2}$

2. ${{n}_{2}}^{2}{{d}_{2}}^{3}={{n}_{1}}^{2}{{d}_{1}}^{3}$

3. ${n}_{1}{{d}_{1}}^{2}={n}_{2}{{d}_{2}}^{2}$

4. ${{n}_{1}}^{2}{d}_{1}={{n}_{2}}^{2}{d}_{2}$

Subtopic: Kepler's Laws |

66%

From NCERT

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