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Unformatted text preview: Force and acceleration We had this problem in the homework A car of mass 2180 kg slows down as the brakes are applied What force is acting to slow the car down? Note that the plot of speed vs time has a uniform slope If theres an acceleration then there must be a force causing that acceleration, and the force F=ma F = ma = (2180 kg )(0.67 m / s ) = 1453 N a = v t = 20 m / s m / s 30 s = 0.67 m / s 2 Oblateness of the Earth In the last lecture, we discussed that gravity is responsible for the planets being round But theyre not exactly round due to the rotational motion of the Earth The Earth tends to be somewhat thicker at the equator than at the poles Not by much 12756 km at equator 12714 km at the poles So a round sphere is still a very good representation of the shape of the Earth How different is the force of gravity at the North Pole and at the Equator? Take a mass of 50 kg. F = G m 1 m 2 d 2 Oblateness of the Earth The Earth tends to be somewhat thicker at the equator than at the poles Not by much 12756 km at equator 12714 km at the poles How different is the force of gravity at the North Pole and at the Equator? Take a mass of 50 kg. F = G m 1 m 2 d 2 How much did the mass change? F equator = W equator = G mm Earth d 2 F equator = (6.67 X 10 11 Nm 2 / kg 2 ) (50 kg )(6 X 10 24 kg ) (6.378 X 10 6 m ) 2 F equator = 491.9 N F pole = (6.67 X 10 11 Nm 2 / kg 2 ) (50 kg )(6 X 10 24 kg ) (6.357 X 10 6 m ) 2 F pole = 495.2 N Jupiter Jupiter is even more oblate since its much larger than the Earth it rotates much faster (one day = 10 hours) its composed mostly of fluid How much would this 50 kg person weigh on the equator of the surface of Jupiter? Jupiter is 300 times as massive as the Earth Why isnt the weight 300 times as much? F = W = G mm Jupiter d 2 F = (6.67 X 10 11 Nm 2 / kg 2 ) (50 kg )(1.90 X 10 27 kg ) (7.13 X 10 7 m ) 2 F = 1246 N Neutron star A neutron star has a mass of 4 X 10 30 kg (about twice the suns mass) and a radius of 10 km (about 1/70000 th that of the sun) What would be the weight of a 50 kg person on the surface of this neutron star?...
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This note was uploaded on 10/17/2011 for the course ECON 101 taught by Professor Thompson during the Spring '11 term at Michigan State University.
 Spring '11
 Thompson

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