**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.

*c*

_{e}2

*Πr*

_{e}= 4.00×10

^{7}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 = |

^{3}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 7.44×10

^{3}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

^{o }below the equator? Again, calculate the required eastwards speed.

^{o }northwards of due east. At this latitude, the radius of the earth is given by

*r*

_{e}cos(38

^{ o }) = 5.03×10

^{7}m and hence the tangential speed is= 366 m/s. But the eastwards speed required by the shuttle is 7.91×10

^{3}cos(38

^{ o }) = 6.23×10

^{3}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 = 5.87×10

^{3}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.

*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

^{-11}sin

*θ*.

*F*= 0.5×10

^{-11}Newtons. The force due to the masses is given by the Universal Law of Gravitation:

F == 4000G |

*G*= 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}= 7.35×10

^{22}and its radius

*r*

_{m}= 1700 kilometers, the earth-moon distance is

*r*= 3.84×10

^{8}meters, the mass of the earth is 5.98×10

^{24}kilograms).

*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 5.98×10

^{24}kilograms.

*x*= 1/2

*gt*

^{2}âá’

*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.