Problems for Kepler's Third Law
Problem :
A geostationary satellite remains at the same position above the
earth's surface at all times. What height above the surface must it be
(assume the orbit is circular)? (Hint: What must the period be?) ($M_e = 5.98
\times 10^{24}$ kilograms and $r_e = 6.38 \times 10^6$ meters).
The period of such a satellite must be 24 hours, since it must rotate
at exactly the same rate as the earth if it is to remain in a constant
position above the surface. 24 hours is 86400 seconds. We can now use
Kepler's Third Law to determine the radius of the orbit:
\begin{equation} r = \left( \frac{GM_eT^2}{4\pi^2} \right)^{1/3} =
\left( \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times
(86400)^2}{4\pi^2} \right)^{1/3} = 4.23 \times 10^7 \rm{meters}
\end{equation}
Now it remains to subtract off the radius of the earth: $4.23 \times
10^7 - 6.38 \times 10^6 = 3.59 \times 10^7$ metres, or roughly 35 000
kilometers.
Problem :
If an ICBM is launched at Washington DC from China, how long will it
take to arrive? The ICMB only fires its rockets at take-off and re-entry and is
allowed to coast in an elliptical orbit reaching a maximum height of 2500
kilometers above the earth. The mass of the earth is $M_e = 5.98 \times
10^{24}$ kg and it has radius $r_e = 6.38 \times 10^6$ m. The mass of the ICBM
is $m = 1000$ kilograms.
The time to impact is going to be half the period given by Kepler's
Third Law.
\begin{equation} T^2 = \frac{4\pi^2 a^3}{GM_e} = \frac{4\pi (6.38
\times 10^6 + 2.5 \times 10^6)^3}{6.67 \times 10^{-11} \times 5.98
\times 10^{24}} = 2.21 \times 10^7 \rm{secs}^2\end{equation}
Taking the square root, and then dividing by two this is approximately
4160 secs, or 69.4 minutes.
Problem :
A satellite has a polar orbit about the earth with a radius of $5.49
\times 10^6$ meters above the earth's surface. On one orbit it passes
over London. Approximately where will it pass over exactly one orbit
later? The radius of the earth is $r_e = 6.38 \times 10^{6}$ meters
and its mass is $5.98 \times 10^{24}$ kilograms.
Let us first calculate the period of the satellite:
\begin{equation} T^2 = \frac{4\pi^2(6.38 \times 10^6 + 5.49 \times
10^6)^3}{6.67 \times 10^{-11} \times 5.98 \times 10^{24}} = 1.83 \times
10^9 \end{equation}
Taking the square root, the period is $1.35 \times 10^4$ secs, which is
3.76 hours. The earth rotates through 360$^{\circ}$ in 24 hours and
thus through 15$^{\circ}$ every hour. Therefore, while the satellite
completes one orbit, the earth has rotated by $3.76 \times 15^{\circ} =
56.4^{\circ}$. Now the sun moves across the sky from east to west, so
the earth must be rotating oppositely, from west towards the east, so
London will have moved East and the satellite will be above a place of
the same latitude, but 56.4$^{\circ}$ west of London.
Problem :
What is the period of a low-earth orbit (that is when $r \approx r_e$)?
Extra hard part: If a hole was dug right through the center of the
earth and a mass was dropped down it, what would occur? Compare this
to the above situation. ($r_e = 6.38 \times 10^6$ meters and $M_e =
5.98 \times 10^{24}$ kilograms and assume that the mass of the earth is
evenly distributed through its volume).
The period of a low-earth orbit can be found directly from Kepler's
Third Law:
\begin{equation}
T = \sqrt{\frac{4\pi^2r_e^3}{GM_e}} = 5070 \rm{secs}
\end{equation}
This is roughly equal to 84.5 minutes. In the case of the hole, we
must consider the force due to the mass of the earth as the object
descends into the earth. Since only matter contained within a radius
less than or equal to where the object is will exert a gravitational
force on it, the force will be a function of the radius. The mass
contained within a sphere of radius r will be given by $M =
\frac{r^3M_e}{r_e^3} = (2.30 \times 10^4)r$. The force then given by
the Universal Law of Gravitation is:
\begin{eqnarray}
F &=& \frac{G(2.30 \times 10^4)r^3m}{r^2} \\
&=& (2.30 \times 10^4)Gmr
\end{eqnarray}
But now we see the force is linearly dependent on $r$ only, times a
constant. This is just like a harmonic oscillator force. Thus the mass dropped into the hole in the
earth will never come out the other side! It will oscillate
harmonically inside the earth. The period of this oscillation =
\begin{equation} 1/T = f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} =
\frac{1}{2\pi}\sqrt{(2.30 \times 10^4)Gm/m} = 1.97 \times 10^{-4}
\rm{Hz} \end{equation}
Hence the period is the inverse of this $T = 5070$ secs. Surprisingly,
this is exactly the same result we derived for the low-earth orbital period.
Problem :
Consider a binary star system with two stars orbiting about their common center
of mass. Each star completes its orbit in 10.2 days. One has mass $50 \times
10^{30}$ kilograms and the other mass $20 \times 10^{30}$ kilograms. Assuming
circular orbit, what is the distance between the stars (the stars will be on
opposite sides of their center of mass, so the distance between them will be
given by the sum of the orbital radii)?
We can imagine we are moving around in the reference
frame
of one of the stars. In this frame, one star is stationary and the other is
moving around it. The radius of this orbit is just the distance between the
stars, $d$, which we are trying to discover. In this case, we want to use the
reduced mass of the system $\mu =
\frac{m_1m_2}{m_1 +
m_2}$, where $m_1$ and $m_2$ are the masses of each star. In this case
Kepler's Third Law becomes:
\begin{equation} T^2 = \frac{4\pi^2 d^3}{G(m_1 + m_2)} \Rightarrow d^3
= \frac{(10.2 \times 24 \times 3600)^2 \times 6.67 \times 10^{-11}
\times 7 \times 10^{31}}{4 pi^2} = 9.19 \times 10^{31} \end{equation}
Hence $d = 4.51 \times 10^{11}$ meters.
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