According to Kepler’s Law of orbits:
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The total energy of a circularly orbiting satellite is
According to Kepler’s Law of periods, The _______________ of the time period of revolution of a planet is proportional to the cube of the ___________ of the ellipse traced out by the planet
A ‘central’ force is always directed
The acceleration due to gravity at the North Pole of Neptune is approximately 10.7 m/s^{2}. Neptune has mass 1.0 x 10^{26} kg and radius 2.5 x 10^{4} km and rotates once around its axis in about 16 h. What is the gravitational force on a 5.0kg object at the north pole of Neptune?
To find the resultant gravitational force acting on the particle m due to a number of masses we need to use:
The acceleration due to gravity at the North Pole of Neptune is approximately 10.7 x m/s^{2}. Neptune has mass 1.0 x 10^{26} kg and radius 2.5 x 10^{4} km and rotates once around its axis in about 16 h. What is the apparent weight a 5.0kg object at Neptune’s equator?
The force of attraction between a hollow spherical shell of uniform density and a point mass situated outside is just as if the entire mass of the shell is
The force of attraction due to a hollow spherical shell of uniform density, on a point mass situated inside it, is
Suppose there existed a planet that went around the sun twice as fast as the earth. What would be its orbital size as compared to that of the earth?
A particle of mass 3m is located 1.00 m from a particle of mass m. Where should you put a third mass M so that the net gravitational force on M due to the two masses is exactly zero?
Moon has a mass of 7.36 x 10^{22} kg, and a radius of 1.74 x 10^{6} m. Calculate the acceleration due to gravity on the moon.
If a film of width l is stretched in the longitudinal direction a distance d by force F, surface tension is given by
Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 1/1700 the mass of the earth. What is the acceleration due to gravity at the surface of Titania
Data: G = 6.67x10^{−11} N m^{2}/kg, RE = 6.38 x 10^{6} m, mE = 5.97 x 10^{24} kg
Which of the following statements correct?
How will you ‘weigh the sun’, that is estimate its mass? The mean orbital radius of the earth around the sun is 1.5 x 10^{8} km.
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