Let the speed of the planet at the perihelion \(P\) in figure shown below be \(v_{_P}\) and the Sun-planet distance \(\mathrm{SP}\) be \(r_{_P}.\) Relation between \((r_{_P},~v_{_P})\) to the corresponding quantities at the aphelion \((r_{_A},~v_{_A})\) is:

              
1. \(v_{_P} r_{_P} =v_{_A} r_{_A}\)
2. \(v_{_A} r_{_P} =v_{_P} r_{_A}\)
3. \(v_{_A} v_{_P} = r_{_A}r_{_P}\)
4. none of these

Subtopic:  Kepler's Laws |
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​Let the speed of the planet at the perihelion \(P\) in the figure shown below be \(v_p\) and the Sun-planet distance \(SP\) be \(r_p\). Will the planet take equal time to traverse \(BAC\) and \(CPB?\)

   

1. no
2. yes
3. depends on the mass of the planet 
4. we can't say anything

Subtopic:  Kepler's Laws |
 58%
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Three equal masses of \(m\) kg each are fixed at the vertices of an equilateral triangle \(ABC.\) What is the force acting on a mass \(2m\) placed at the centroid \(G\) of the triangle
(
Take \(AG=BG=CG=1\) m.)

       

1. \(Gm^2(\hat{i}+\hat{j})\)
2. \(Gm^2(\hat{i}-\hat{j})\)
3. zero
4. \(2Gm^2(\hat{i}+\hat{j})\)

Subtopic:  Newton's Law of Gravitation |
 82%
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Three equal masses of m kg each are fixed at the vertices of an equilateral triangle ABC. What is the force acting on a mass 2m placed at the centroid G of the triangle if the mass at the vertex A is doubled? Take AG = BG = CG = 1 m.

1.   Gm2 i^+j^

2.   Gm2 i^-j^

3.   0

4.   2Gm2 j^

Subtopic:  Newton's Law of Gravitation |
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The potential energy of a system of four particles placed at the vertices of a square of side l (as shown in the figure below) and the potential at the centre of the square, respectively, are:

1.   -5.41Gm2l and 0

2.   0 and -5.41Gm2l 

3.   -5.41Gm2l and -42Gml

4.   0 and 0

Subtopic:  Gravitational Potential Energy |
 58%
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Two uniform solid spheres of equal radii \({R},\) but mass \({M}\) and \(4M\) have a centre to centre separation \(6R,\) as shown in the figure. The two spheres are held fixed. A projectile of mass \(m\) is projected from the surface of the sphere of mass \(M\) directly towards the centre of the second sphere. The expression for the minimum speed \(v\) of the projectile so that it reaches the surface of the second sphere is:

      
1. \(\left(\frac{3 {GM}}{5 {R}}\right)^{1 / 2}\)
2. \(\left(\frac{2 {GM}}{5 {R}}\right)^{1 / 2}\)
3. \(\left(\frac{3 {GM}}{2 {R}}\right)^{1 / 2}\)
4. \(\left(\frac{5 {GM}}{3 {R}}\right)^{1 / 2}\)

Subtopic:  Gravitational Potential Energy |
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The planet Mars has two moons, Phobos and Delmos. Phobos has a period of \(7\) hours, \(39\) minutes and an orbital radius of 9.4×103 km. The mass of mars is:
1. 6.48×1023 kg
2. 6.48×1025 kg
3. 6.48×1020 kg
4. 6.48×1021 kg

Subtopic:  Satellite |
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You are given the following data: g 9.81 m/s2, RE = 6.37×106 m, the distance to the moon, R = 3.84×108 m and the time period of the moon’s revolution is 27.3 days. Mass of the Earth ME in two different ways is:

1.   5.97×1024 kg and 6.02×1024 kg

2.   5.97×1024 kg and 6.02×1023 kg

3.   5.97×1023 kg and 6.02×1024 kg

4.   5.97×1023 kg and 6.02×1023 kg

Subtopic:  Satellite |
 55%
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Constant k = 10-13 s2m-3 in days and kilometres is?

1.   10-13 d2km-3

2.   1.33×1014 dkm-3

3.   10-13 d2km

4.   1.33×10-14 d2km-3

Subtopic:  Satellite |
 57%
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The moon is at a distance of \(3.84\times10^5~\text{km}\) from the earth. Its time period of revolution in days is: \((\text{Given }k=\frac{4\pi^2}{GM_E}=1.33\times10^{-14}~\text{days}^{2}-\text{km}^{-3})\)
1. \(17.3\) days
2. \(33.7\) days
3. \(27.3\) days
4. \(4\) days
Subtopic:  Satellite |
 63%
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