A 400 kg satellite is in a circular orbit of radius 2RE about the Earth. What are the changes in the kinetic and potential energies respectively to transfer it to a circular orbit of radius 4RE. (where RE is the radius of the earth)

1. 3.13×109 J and 6.25×109 J

2. 3.13×109 J and -6.25×109 J

3. -3.13×109 J and -6.25×109 J

4. -3.13×108 J and -6.25×108 J

Subtopic:  Satellite |
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Assume that earth and mars move in circular orbits around the sun, with the martian orbit being \(1.52\) times the orbital radius of the earth. The length of the martian year in days is approximately:
(Take \((1.52)^{3/2}=1.87\))

1. \(344\) days 2. \(684\) days
3. \(584\) days 4. \(484\) days
 

Subtopic:  Kepler's Laws |
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A \(400\) kg satellite is in a circular orbit of radius \(2R_E\) (where \(R_E\) is the radius of the earth) about the Earth. How much energy is required to transfer it to a circular orbit of radius \(4R_E\)\(?\)
(Given \(R_E=6.4\times10^{6}\) m)

1. \(3.13\times10^{9}\) J 2. \(3.13\times10^{10}\) J
3. \(4.13\times10^{9}\) J 4. \(4.13\times10^{8}\) J
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:
 \((\text{Given }k=\dfrac{4\pi^2}{GM_E}=1.33\times10^{-14}~\text{days}^{2}-\text{km}^{-3})\)
1. \(17.3\) days
2. \(33.7\) days
3. \(27.3\) days
4. \(4\) days
Subtopic:  Satellite |
 63%
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Constant \(k   =   10^{- 13} ~ \text s^{2}~ \text m^{- 3}\) in days and kilometres is?
1. \(10^{- 13} ~ \text d^{2} ~\text{km}^{- 3}\) 2. \(1 . 33 \times 10^{14}   \text{ dkm}^{- 3}\)
3. \(10^{- 13} ~ \text d^{2} ~\text {km}\) 4. \(1 . 33 \times 10^{- 14} \text{  d}^{2} \text{ km}^{- 3}\)
Subtopic:  Satellite |
 57%
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You are given the following data: \(g = 9.81~ \text{m/s}^{2}\), \(R_{E}   =   6 . 37 \times 10^{6}~\text m\), the distance to the moon, \(R = 3 . 84 \times 10^{8}~\text m\) and the time period of the moon’s revolution is 27.3 days. Mass of the Earth \(M_{E}\) in two different ways is:

1. \(5 . 97 \times 10^{24}  ~ \text{kg and }6 . 02 \times 10^{24}   \text{ kg}\)

2. \(5 . 97 \times 10^{24}  \text{ kg and }  6 . 02 \times 10^{23}  \text{ kg}\)

3. \(5 . 97 \times 10^{23}  ~ \text{kg and }6 . 02 \times 10^{24}   \text{ kg}\)

4. \(5 . 97 \times 10^{23}  \text{ kg and }  6 . 02 \times 10^{23}  \text{ kg}\)

Subtopic:  Satellite |
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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 \(9 . 4 \times 10^{3}\) km. The mass of mars is:

1. \(6 . 48 \times 10^{23}  \text{ kg}\) 2. \(6 . 48 \times 10^{25}  \text{ kg}\)
3. \(6 . 48 \times 10^{20}  \text{ kg}\) 4. \(6 . 48 \times 10^{21}  \text{ kg}\)
Subtopic:  Satellite |
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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. \(\left(\dfrac{3 {GM}}{5 {R}}\right)^{1 / 2}\) 2. \(\left(\dfrac{2 {GM}}{5 {R}}\right)^{1 / 2}\)
3. \(\left(\dfrac{3 {GM}}{2 {R}}\right)^{1 / 2}\) 4. \(\left(\dfrac{5 {GM}}{3 {R}}\right)^{1 / 2}\)
Subtopic:  Gravitational Potential Energy |
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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. \(- 5 . 41 \dfrac{Gm^{2}}{l}\) and 0

2. 0 and \(- 5 . 41 \dfrac{Gm^{2}}{l}\) 

3. \(- 5 . 41 \dfrac{Gm^{2}}{l}\) and \(- 4 \sqrt{2} \dfrac{Gm}{l}\)

4. 0 and 0

Subtopic:  Gravitational Potential Energy |
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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.   \(Gm^{2}   \left(\hat{i} + \hat{j}\right)\)

2.   \(Gm^{2}   \left(\hat{i} - \hat{j}\right)\)

3.   0

4.   \(2Gm^{2} \hat{j}\)

Subtopic:  Newton's Law of Gravitation |
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