Satellite in orbit 2

Problem statement:

Scientists want to place a {msat} kg satellite in orbit around Mars. They plan to have the satellite orbit at a speed of {vsat} m/s in a perfectly circular orbit. Here is some information that may help solve this problem:

mass of mars: $M$ = 6.4191 × 10²³ kg
radius of orbit: $R$= 3.397 × 10⁶ m
$G$ = 6.67428 × 10^-11 N-m²/kg²

• (1) What is the radius of the orbit? We are given the speed and want the radius of the orbit. The principle is to set the gravitational force equal to the centripetal force: $$ \frac{GMm}{R^2} = \frac{mv^2}{R}$$ Where $M$ is the mass of mars and $m$ the mass of the satellite. Notice that $m$ cancels.
  Solve for $R$: $$R = \frac{GM}{v^2}$$
  This deserves a unit check. The units of $G$ are [N m²/kg²] × [kg]/[m²/s²]
  Reduces to [kg m/s² m²/kg²] × [kg]/[m²/s²] = [m] (It really does!)
  Just plug in the number for the orbital speed to get the answer.
  
  
• (5)What should the speed of the orbit be if we want the satellite to take 8 times longer to complete one full revolution of its orbit?

  This is the part that causes the greatest headache. You can save a lot of trouble by using ratios. We want a ratio involving $v$ and $T$ equal to a constant with no other variable. The problem is that $v$, $R$ an $T$ are all related. $$ \frac{GMm}{R^2} = \frac{mv^2}{R}$$ Get rid of the $v$ and introduce the $T$ (the period) by using the old standby $$ v =\frac{2\pi R}{T}$$ sub into the previous eqn and get rid of the $m$'s $$ \frac{GM}{R^2} = \frac{(\frac{2\pi R}{T})^2}{R}$$ We get our ratio as follows: $$\frac{R^3}{T^2} = \frac{GM}{4\pi^2}$$ (This is Kepler's 3rd law.) Darn. We've got $R$ in the ratio not $v$. (This is actually what we need for the previous problem.

  Anyhow, Substitute $R = Tv/2\pi$ $$\frac{(Tv/2\pi)^3}{T^2} = \frac{GM}{4\pi^2}$$ $$v^3 T = {\rm constant}$$ The important point is that this ratio is equal to a constant and we really don't care what that constant is. If we're given that $T_2/T_1 = 8$ we can easily get $v_2/v_1$. $$\left(\frac{v_2}{v_1} \right)^3 = \frac{T_1}{T_2} $$ $$\frac{v_2}{v_1} = \sqrt[3]{ \frac{T_1}{T_2} } = \left( \frac{T_1}{T_2}\right)^{1/3}$$ Thus $$\frac{v_2}{v_1} = \frac{1}{2}$$