“ every particle of matter in the universe attracts every other particle with a force which is directly proportional to the product of their masses, and inversely proportional to the square of their distance apart ”
The units of gravitational field strength are N kg -1
The gravitational field strength is a vector quantity whose direction is given by the direction of the force a mass would experience if placed at the point of interest
If we replace the particle M with a sphere of mass M and radius R then relying on the fact that the sphere behaves as a point mass situated at its centre the field strength at the surface of the sphere will be given by
But the field strength is equal to the acceleration that is produced on the mass, hence we have that the acceleration of free fall at the surface of the Earth, g
The potential is therefore a measure of the amount of work that has to be done to move particles between points in a gravitational field and its units are J kg –1
The work done is independent of the path taken between the two points in the field, as it is the difference between the initial and final potentials that give the value
The gravitational field strength is a vector quantity whose direction is given by the direction of the force a mass would experience if placed at the point of interest
A 2000kg spacecraft in orbit at R above the Earth of radius R. The potential at the Earth's surface is –60 MJ kg -1 . What is the change in potential energy if the spacecraft returns to Earth
26.
Graphs Gravitational field strength versus distance g α 1/r 2 Gravitational potential versus distance V α -1/r
In some situations the use of the conservation of energy can be a much simpler method than using the kinematic equations
Solving projectile motion problems makes use of the fact that E k + E p = constant at every point in the objects flight (assuming no loss of energy due to friction)
A ball is projected at 25.0 ms -1 at an angle of 40.0 0 to the horizontal. The ball is released 2.00m above the ground. Taking g = 10.0 ms -2 . Find the maximum height it reaches.
42.
Solution 2.0m 25.0 ms -1 A B v = v horizontal H
The square of the times of revolution of the planets (i.e. Their periodic time T ) about the sun are proportional to the cubes of their mean distance ( r ) from it.
We have considered a circular orbit but more advanced mathematics gives the same result for an elliptical one.
i.e. The balance registers the weight of the body as zero
It is usual to refer to a body in this situation as being weightless
The term should be used with care, a gravitational pull of magnitude mg acts on the body whether it is in free fall or not, and therefore, in the strictest sense it has weight even when in free fall.
The reason it is said to be weightless is that, whilst falling freely, it exerts no force on its support.
Similarly, a man standing on the floor of a lift would exert no force on the floor if the lift were in free fall. In accordance with Newton´s third law, the floor of the lift would exert no upwards push on the man and therefore he would not have the sensation of weight.
An astronaut in an orbiting spacecraft has a centripetal acceleration equal to g 1 , where g 1 is the acceleration due to gravity at the height of the orbit
The spacecraft also has the same centripetal acceleration
The astronaut therefore has no acceleration relative to his spacecraft, i.e. he is weightless
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