Download
1 / 34
340 likes | 552 Views
Fields and Forces. Gravitational Force Field. 6.1.1 State Newton’s universal law of gravitation.
E N D
Gravitational Force Field 6.1.1 State Newton’s universal law of gravitation “Every material particle in the Universe attracts every other material particle with a force that is directly proportional to the product of the masses of the particles and that is inversely proportional to the square of the distance between them.” (Newton’s Principia)
6.1.2 Define gravitational field strength 6.1.4 Derive an expression for gravitational field strength at the surface of a planet, assuming that all its mass is concentrated at its centre.
6.1.3 Determine the gravitational field due to one or more point masses
6.1.5 Solve problems involving gravitational forces and fields.
Escape Velocity If the kinetic energy of an object launched from the Earth were equal in magnitude to the potential energy, then in the absence of friction resistance it could escape from the Earth. Equate expressions for the two energies to find the escape velocity. kinetic energy potential energy From http://hyperphysics.phy-astr.gsu.edu/hbase/vesc.html
Orbit Velocity Find the orbit velocity for a circular orbit by setting the gravitational force equal to the required centripetal force.
Kepler’s Laws Show a derivation for this.
Astronauts in orbit do not experience true weightlessness. True weightlessness occurs only in deep space, out of reach of the gravitational fields of stars and planets, where a person’s weight is as close to zero as it can be.
Suppose a person of mass ‘m’ is standing on a weighing machine placed in a lift/elevator. The actual weight of the man is ‘mg’. This weight presses the machine. The machine also exerts a reactionary force ‘R’ on the person in upward direction where ‘R = W’ (Newton’s Third Law). Thus, two forces are acting on the person; gravity force ‘mg’ and reactionary force ‘R’. Since the two forces are in opposite directions, the net force on the person is given by: F = mg – R (downwards) But the person is at rest (no acceleration) hence the net force on him should be zero. F = mg – R = 0R = mg R = WW = mg That is, that apparent weight is equal to the actual weight of the person. From http://www.xamplified.com/apparent-weight/
Now, suppose the elevator is moving uniformly in the upward directions. The acceleration of the lift be ‘a’. Again, the net force on the person is F = mg – R (downwards). according to Newton’s Second Law of Motion, the net force on him is given by F = ma (upwards) or F = -ma (downwards). -ma = mg – R R = mg + ma R = w w = m(g + a) Thus, in this condition, the apparent weight ‘W’ of the person is greater than the gravity force ‘mg’. That is, the person will experience his weight to be increased than in the normal state.
Now, suppose the lift is coming down with acceleration ‘a’ (fig (c)). Again, the net force on the person is F = mg – R (downwards) As the acceleration ‘a’ is directed downwards, hence the net force F = ma should also be directed downwards. -ma = mg – R R = mg + ma R = W W = mg – ma W = m(g – a) In this situation, the person feels his apparent weight to be decreased than the usual weight.
If the string of the descending lift is broken then it will fall freely under gravity. In this situation, a = g Then from above equation W = mg – maW = mg – mg = 0 Now the machine will read zero or the body becomes weightless. In this state, the person will not feel even the weight of a body placed on his head. If the person drops a ball from his hand, the ball will not fall down on the floor of the lift, but will appear floating in air. This is because the ball and the lift both are falling under the same acceleration g. If the acceleration of the descending lift is greater than the acceleration due to gravity ‘g’ then W = mg – ma The apparent weight of the person will be negative. Under this condition, the person will rise from the floor of the lift and stick to the ceiling of the lift.
