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# Kepler and Gravitation

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#### 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: $$r = \left( \frac{GM_eT^2}{4\pi^2} \right)^{1/3} = \left( \frac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (86400)^2}{4\pi^2} \right)^{1/3} = 4.23 \times 10^7 \rm{meters}$$ 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. $$T^2 = \frac{4\pi^2 a^3}{GM_e} = \frac{4\pi (6.38 \times 10^6 + 2.5 \times 10^6)^3}{6.67 \times 10^{-11} \times 5.98 \times 10^{24}} = 2.21 \times 10^7 \rm{secs}^2$$ 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: $$T^2 = \frac{4\pi^2(6.38 \times 10^6 + 5.49 \times 10^6)^3}{6.67 \times 10^{-11} \times 5.98 \times 10^{24}} = 1.83 \times 10^9$$ 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: $$T = \sqrt{\frac{4\pi^2r_e^3}{GM_e}} = 5070 \rm{secs}$$ 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 = $$1/T = f = \frac{1}{2\pi}\sqrt{\frac{k}{m}} = \frac{1}{2\pi}\sqrt{(2.30 \times 10^4)Gm/m} = 1.97 \times 10^{-4} \rm{Hz}$$ 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: $$T^2 = \frac{4\pi^2 d^3}{G(m_1 + m_2)} \Rightarrow d^3 = \frac{(10.2 \times 24 \times 3600)^2 \times 6.67 \times 10^{-11} \times 7 \times 10^{31}}{4 pi^2} = 9.19 \times 10^{31}$$ Hence $d = 4.51 \times 10^{11}$ meters.