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Kepler's Law of Orbits:
The escape velocity is the minimum velocity required to leave a planet or moon. For a rocket or other object to leave a planet, it must overcome the pull of gravity.
V_{escape }= √2GM/R
V_{escape }= 11184 m/sec approximate to 11.2 km/sec
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
Kepler's 3rd law is a mathematical formula. It means that if you know the period of a planet's orbit (T = how long it takes the planet to go around the Sun), then you can determine that planet's distance from the Sun (a is the length of the semimajor axis of the planet's orbit)
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?
Gravitational Force:
F = mg
⇒ F = 5×10.7 = 53.5 N
To find the resultant gravitational force acting on the particle m due to a number of masses we need to use:
According to properties of gravitational force, gravitational force between the particles is independent of the presence or absence of other particles; so the principle of superposition is valid i.e. force on a particle due to number of particles is the resultant of forces due to individual particles.
The satellite which appear stationary relative to earth, such satellites are called geostationary satellites and will have a period of rotation as the period of rotation of earth i.e 24 Hrs.
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?
Time period of revolution of earth around sun T_{e}= 1 Year
Time period of revolution of planet around sun,T_{p}=0.5 Year
Orbital size of earth, r_{e}= 1 A.U
Orbital size of the planet,r_{p}=?
Applying Kepler's third law we get:
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?
Let x be the distance of third particle from the mass 3m. then (1  x) is the distance will be the distance from mass m. If we require that net force on another mass M be zero, then we must have the following :
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
We know that gravitational acceleration, g = GM/R^{2}
We know that M = Mass of earth /1700 and R = Radius of earth /8
Hence we get g = 64/1700 times the gravitational acceleration of earth
I.e. g = 64/1700 x 9.8
= 0.37 m/s^{2}
The acceleration due to gravity due to a body at point P on the surface of earth is
Let the body is placed to point Q at a height h abovr the surface of earth, then acceleration due to gravity:
From above expression we can clearly say that with increase in altitiude h , the value of acceleration due to gravity decreases.
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.
R = Radius of Orbit of earth = 1.5 x 10^{8} km = 1.5 x 10^{11}m
T = time Period of earth around the sun = 365 Days
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