Newtons Laws EX 36

The shuttle was placed into a circular orbit of radius $r=(6.37\times 10^{6})+(3.15\times 10^{5})=6.685\times 10^{6}$ while the satellite was in the orbit of radius $r=(6.37\times 10^{6})+(3.60\times 10^{5})=6.73\times 10^{6}$. The shuttle was initially on the opposite side of the satellite and we want to know how long will it take to be beneath the satellite. We need to find the period for both of these satellites. To get this we will use the formula for the velocity.

$v={2\pi r \over T}$

Substitute the formula for the velocity into the expression for the Newton's second law:

${GM_{E}M_{S} \over r^{2}}=M_{S}{v^{2} \over r}$

${GM_{E}M_{S} \over r^{2}}=M_{S}{({2\pi r \over T})^{2} \over r}$

${GM_{E} \over r^{3}}={4\pi ^{2} \over T^{2}}$

${(6.67\times 10^{-11})(6\times 10^{24}) \over (6.685\times 10^{6})^{3}}={4\pi ^{2} \over T^{2}}$

and find the period of the shuttle to be $5458s$.

Similarly, for Westar, we obtain the equation

${(6.67\times 10^{-11})(6\times 10^{24}) \over (6.73\times 10^{6})^{3}}={4\pi ^{2} \over T^{2}}$

and solve it to find the period of the Westar to be $5499s$.

Since at the beginning the two satellites were on the opposite side of the Earth, the shuttle has to make more revolutions than the satellite. Let's assume that in some time $\Delta t$ the shuttle is beneath the satellite. Given the period of the shuttle ($5458s$), we deduce that the number of revolutions made by the shuttle is given by

$N={\Delta t \over T}={\Delta t \over 5458}$

It has moved through an angle of

$\Delta \theta _{S}=2\pi N=2\pi ({\Delta t \over 5458})$

During the same time the Westar moved through an angle

$\Delta \theta _{W}=2\pi ({\Delta t \over 5499})$

The shuttle needs to "catch up" with Westar. Therefore, the angle $\Delta \theta _{S}$ is larger than $\Delta \theta _{W}$. Since they were initially on the opposite side of Earth:

$\Delta \theta _{S}=\Delta \theta _{W}+\pi$

or

$2\pi ({\Delta t \over 5458})=2\pi ({\Delta t \over 5499})+\pi$

Solving for $\Delta t$ we find $\Delta t=3.66\times 10^{5}s$.