motion of satellites

1304 The Motion of Satellites

Satellites: Circular orbits

By the law of gravitation, the net force on the satellite of mass m has magnitude Fg=GmEm/r2F_g = Gm_Em/r^2, and is in the same direction as the acceleration. Newton's second law then tells us that

Gmemr2=mv2r(1)\begin{aligned} \frac{Gm_e m}{r^2} = \frac{mv^2}{r} \quad &\text{(1)} \end{aligned}

Solving this for vv, we finding

v=GmEr(2)\begin{aligned} v = \sqrt{\frac{Gm_E}{r}} \quad &\text{(2)} \end{aligned}

The speed vv is the distance 2πr2\pi r traveld in one revolution, divided by the period

v=2πrT⇒T=2πrv\begin{aligned} v &= \frac{2\pi r}{T}\\ \To T&= \frac{2\pi r}{v} \end{aligned}

The total mechanical energy can be determined as follows:

E=K+U=12mv2+(−GmEmr)=12m(GmEmr)+(−GmEmr)=−GmEm2r(3)\begin{aligned} E &= K + U = \frac{1}{2}mv^2 + (-\frac{Gm_E m}{r})\\ &= \frac{1}{2}m(\frac{Gm_E m}{r}) + (-\frac{Gm_E m}{r})\\ &= -\frac{Gm_E m}{2r} \quad &\text{(3)} \end{aligned}

Exercises

20, 23, 24, 25

(13.20) An earth satellite moves in a circular orbit with an orbital speed of 6200 m/s. Find (a) the time of one revolution of the satellite; (b) the radial acceleration of the satellite in its orbit.

Solution

Let the period TT and the radial acceleration be aa, we first calculate the radius of the orbit rr,

v=GmEr⇒r=GmEv2=G⋅5.97⋅102462002=10358975mT=2πrv=2π⋅103589756200=10498s=175mina=v2r=6200210358975=3.71m/s2\begin{aligned} v &= \sqrt{\frac{Gm_E}{r}}\\ \To r &= \frac{Gm_E}{v^2}\\ &= \frac{G \cdot 5.97 \cdot 10^{24}}{6200^2}\\ &= 10358975m\\ T &= \frac{2\pi r}{v}\\ &= \frac{2\pi \cdot 10358975}{6200} = 10498s = 175min\\ a &= \frac{v^2}{r}\\ &= \frac{6200^2}{10358975} = 3.71m/s^2 \end{aligned}

(13.23) Two satellites are in circular orbits around a planet that has radius 9.00×106m9.00 \times 10^6m. One satellite has mass 68.0 kg, orbital radius 7.00×107m7.00 \times 10^7 m, and orbital speed 4800 m/s. The second sat ellite has mass 84.0 kg and orbital radius 3.00×107m3.00 \times 10^7 m. What is the orbital speed of this second satellite?

Solution

Todo
v2=7332m/sv_2 = 7332m/s

(13.24) International Space Station. In its orbit each day, the International Space Station makes 15.65 revolutions around the earth. Assuming a circular orbit, how high is this satellite above the surface of the earth?

Solution

T=24⋅360015.65=5520.8sv=2πrT=Gmer⇒r=GM⋅(T2π)23=6.67⋅10−11⋅5.97⋅1024⋅(5520.82π)23=6749108.5mh=r−RE=6749108.5−6.37⋅106=379108.5m≈380km\begin{aligned} T &= \frac{24 \cdot 3600}{15.65} = 5520.8s\\ v &= \sqrt{2\pi r}{T} = \sqrt{\frac{Gm_e}{r}}\\ \To r &= \sqrt[3]{GM\cdot\left(\frac{T}{2\pi}\right)^2}\\ &= \sqrt[3]{6.67\cdot 10^{-11} \cdot 5.97 \cdot 10^{24}\cdot\left(\frac{5520.8}{2\pi}\right)^2}\\ &= 6749108.5m\\ h & = r - R_E \\ &= 6749108.5 - 6.37 \cdot 10^6 = 379108.5m \approx 380km \end{aligned}

The hight above the surface earth is approximately 380km.

(13.25) Deimos, a moon of Mars, is about 12 km in diameter with mass 1.5×1015kg1.5 \times 10^{15} kg. Suppose you are stranded alone on Deimos and want to play a one-person game of baseball. You would be the pitcher, and you would be the batter!
(a) With what speed would you have to throw a baseball so that it would go into a circular orbit just above the surface and return to you so you could hit it? Do you think you could actually throw it at this speed?
(b) How long (in hours) after throwing the ball should you be ready to hit it? Would this be an action-packed baseball game?

Solution

v=GmR=6.67⋅10−11⋅1.5⋅10156000=4.1m/sT=2πRv=2π⋅60004.1≈2.6h\begin{aligned} v &= \sqrt{\frac{Gm}{R}}\\ &= \sqrt{\frac{6.67\cdot 10^{-11} \cdot 1.5 \cdot 10^{15}}{6000}} = 4.1m/s\\ T &= \frac{2\pi R}{v}\\ &= \frac{2\pi \cdot 6000}{4.1} \approx 2.6h \end{aligned}