Worked examples · Sheet 3

Satellites in circular orbit

Theory used: Speed and period in a circular orbit Geostationary satellite Kepler's third law for circular orbits Gravitational acceleration at the surface

Go to: Exercise 1 Exercise 2 Exercise 3

Exercise 1

A satellite travels on a circular orbit at an altitude $h = 600 km$ above the Earth’s surface. Compute the speed of the satellite and the period of the orbit. Data: $G = 6.67 \cdot 1 0^{- 11} N \cdot m^{2} /k g^{2}$ , $M_{T} = 5.97 \cdot 1 0^{24} kg$ , $R_{T} = 6.37 \cdot 1 0^{6} m$ ; atmospheric drag and the attraction of the Moon and the Sun are neglected.

r = R_{T} + h, \frac{G M _{T} m}{r ^{2}} = m \frac{v ^{2}}{r} ⟹ v = \frac{G M _{T}}{r}, T = \frac{2 π r}{v}

r = 6.37 \cdot 1 0^{6} m + 0.60 \cdot 1 0^{6} m = 6.97 \cdot 1 0^{6} m, G M_{T} = 3.98 \cdot 1 0^{14} m^{3} / s^{2}

v = \frac{3.98 \cdot 1 0 ^{14}}{6.97 \cdot 1 0 ^{6}} m/s = 5.71 \cdot 1 0^{7} m/s = 7.56 \cdot 1 0^{3} m/s

T = \frac{2 π \cdot 6.97 \cdot 1 0 ^{6}}{7.56 \cdot 1 0 ^{3}} s = 5.79 \cdot 1 0^{3} s \approx 96.6 min

The radius of the orbit is the distance from the center of the Earth, so the Earth’s radius must be added to the altitude. The only force on the satellite is the gravitational attraction, and in a circular orbit this force provides the centripetal acceleration; equating the two expressions, the mass $m$ of the satellite cancels and we obtain the speed in a circular orbit. The period is the time needed to travel the circumference of radius $r$ at speed $v$ .

The satellite travels at $7.56 km/s$ and completes one revolution in about an hour and a half. As a check, the speed is slightly less than that of the grazing orbit, $7.91 km/s$ , as it must be for a slightly higher orbit.

Exercise 2

A geostationary satellite travels on an equatorial circular orbit with period equal to the sidereal day, $T = 86164 s$ . Compute the radius of the orbit, the altitude above the surface and the speed of the satellite. Use $G M_{T} = 3.98 \cdot 1 0^{14} m^{3} / s^{2}$ and $R_{T} = 6.37 \cdot 1 0^{6} m$ .

T = 2 π \frac{r ^{3}}{G M _{T}} ⟺ r = 3 \frac{G M _{T} T ^{2}}{4 π ^{2}}

r = 3 \frac{3.98 \cdot 1 0 ^{14} \cdot ( 8.6164 \cdot 1 0 ^{4} ) ^{2}}{4 π ^{2}} m = 3 7.48 \cdot 1 0^{22} m = 4.21 \cdot 1 0^{7} m

h = r - R_{T} = 4.214 \cdot 1 0^{7} m - 0.637 \cdot 1 0^{7} m = 3.58 \cdot 1 0^{7} m

v = \frac{2 π r}{T} = \frac{2 π \cdot 4.214 \cdot 1 0 ^{7}}{86 164} m/s = 3.07 \cdot 1 0^{3} m/s

The geostationary satellite must complete one revolution while the Earth completes one rotation relative to the stars, so its period is the sidereal day, which is about four minutes shorter than the $24$ -hour solar day. From the formula for the period in a circular orbit we isolate $r^{3}$ and take the cube root.

The orbit has radius $4.21 \cdot 1 0^{7} m$ , that is, about $6.6$ Earth radii, and the altitude is about $35800 km$ . The speed is about $3.07 km/s$ , much less than that of a satellite in low orbit. As a check, the same speed is obtained from $\frac{G M _{T}}{r} = 9.45 \cdot 1 0^{6} m/s = 3.07 \cdot 1 0^{3} m/s$ .

Exercise 3

Io, the innermost of the satellites discovered by Galileo, travels around Jupiter on a nearly circular orbit of radius $r = 4.22 \cdot 1 0^{8} m$ with period $T = 1.769$ days. Compute the mass of Jupiter, with $G = 6.67 \cdot 1 0^{- 11} N \cdot m^{2} /k g^{2}$ .

T^{2} = \frac{4 π ^{2}}{G M _{G}} r^{3} ⟺ M_{G} = \frac{4 π ^{2} r ^{3}}{G T ^{2}}

T = 1.769 \cdot 86400 s = 1.528 \cdot 1 0^{5} s

M_{G} = \frac{4 π ^{2} \cdot ( 4.22 \cdot 1 0 ^{8} ) ^{3}}{6.67 \cdot 1 0 ^{- 11} \cdot ( 1.528 \cdot 1 0 ^{5} ) ^{2}} kg = \frac{39.48 \cdot 7.515 \cdot 1 0 ^{25}}{6.67 \cdot 1 0 ^{- 11} \cdot 2.336 \cdot 1 0 ^{10}} kg = 1.90 \cdot 1 0^{27} kg

The motion of Io is determined by the attraction of Jupiter alone, whose mass is much greater than that of the satellite; the perturbations due to the other satellites and to the Sun are neglected. By Kepler’s third law in the case of circular orbits the ratio $\frac{T ^{2}}{r ^{3}}$ depends only on the mass of the central body, so from the period and the radius of the orbit we obtain the mass of Jupiter. The period must be converted to seconds before substituting.

Jupiter has mass $1.90 \cdot 1 0^{27} kg$ , about $320$ times the mass of the Earth. The mass of the satellite does not appear in the calculation, so this method measures the mass of the planets that have at least one observable satellite.