## Monday, 15 September 2014

### NEWTONS LAW OF GRAVITATION PROBLEMS

Problem1 : 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 ( r r e )? 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 c e 2Πr e = 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? = âá’v = Plugging in the radius and mass of the earth yields 7.91×103 m/s. If the shuttle is to be launched into an eastward orbit, the rotation of the earth helps out: the eastward speed required is the speed required for a low earth orbit, minus the speed of rotation. That is 7.44×103 m/s.
Problem2 : What if the space shuttle, instead of being launched from the equator, is launched from Melbourne, Australia, at a latitude of 38 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 38 northwards of due east. At this latitude, the radius of the earth is given by r ecos(38 o ) = 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(38 o ) = 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:6230–366 = 5.87×103 m/s. Of course, the shuttle also needs to be given some velocity in the northwards direction.
Problem 3: Calculate the value for G which would be given by a Cavendish apparatus (in an imaginary universe) set up as shown: Cavendish apparatus.
and with M = 5 kilograms, m = 1 kilogram, θ = 30 degrees, and r = 0.05 meters. Also, the force exerted by the fiber is given by F = 10-11sinθ .
The force on the fiber is given by F = 0.5×10-11 Newtons. The force due to the masses is given by the Universal Law of Gravitation:
 F = = 4000G

Therefore we have 4000G = 0.5×10-11 which implies that G = 1.25×10-15 N m 2/kg 2 .
Problem 4: 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 M m= 7.35×1022 and its radius r m = 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: g e = =  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 = 1/2gt2âá’t = = = 452 secs. In the second case, we must replace r e by r e + 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.