Problem : If a space shuttle is launched from the equator, what eastward speed (relative to the ground) is required to propel it into a low-earth orbit (rre)? Take into account the rotation of the earth in this problem.The earth rotates from west to east. The circumference of the earth is approximately ce 2Πre = 4.00×107 m. The earth rotates through one twenty-fourth of this distance every hour, and the tangential velocity can be calculated to be 464 m/s. What speed is required for a low-earth orbit?
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Problem : What if the space shuttle, instead of being launched from the equator, is launched from Melbourne, Australia, at a latitude of 38o below the equator? Again, calculate the required eastwards speed.The shuttle requires the same velocity to achieve low-earth orbit, but now in a direction 38o northwards of due east. At this latitude, the radius of the earth is given by recos(38o) = 5.03×107 m and hence the tangential speed is = 366 m/s. But the eastwards speed required by the shuttle is 7.91×103cos(38o) = 6.23×103 m/s. Again we can subtract the eastwards speed of the earth, which we calculated for this latitude, and thus we have a required eastwards speed of: 6230366 = 5.87×103 m/s. Of course, the shuttle also needs to be given some velocity in the northwards direction.
Problem : Calculate the value for G which would be given by a Cavendish apparatus (in an imaginary universe) set up as shown:
|F = = 4000G|
Problem : Calculate g on the moon due to the moon's gravity. Compare this to the gravitational force exerted on a mass m on the moon due to the earth (ie. what is g for the earth at the height of the moon?)(The mass of the moon is Mm = 7.35×1022 and its radius rm = 1700 kilometers, the earth-moon distance is r = 3.84×108 meters, the mass of the earth is 5.98×1024 kilograms).The value for g is given by: g = = 1.7. The g value of the earth at the moon distance is given by: ge = = 0.003. The gravity on the moon is therefore approximately 6 times less than that on the earth.
Problem : A ball is dropped from a height of 1000 kilometers above the earth. Calculate the time taken for it to hit the ground as given by i) assuming g takes the value at the surface of the earth; ii) assuming g takes the value at the top of the ball's path. The mass of the earth is 5.98×1024 kilograms.In the first case, we can just use the standard kinematical equation: x = 1/2gt2âát = = = 452 secs. In the second case, we must replace re by re + h: t = = 523 secs. As we would expect, the time is longer, since the gravitational force is weaker further away from the earth, accelerating the ball more slowly.
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