Gravitational Fields: Question 5

Worked solution

Part (a): Conditions for a geostationary orbit

Besides matching Halvane’s rotation period, a geostationary satellite must also:

1. orbit in the plane of Halvane’s equator (so that it stays above the same latitude, namely $0°$); and
2. orbit in the same direction as Halvane’s rotation, i.e. west to east.

(A circular, rather than elliptical, orbit is also required for the satellite to stay at constant radius and constant angular speed above the same point, but the two conditions above are the standard additional requirements alongside a circular orbit.)

Part (b): Deriving the geostationary radius formula

For the satellite (mass $m$) in a circular orbit of radius $r$ with period $T$, its angular speed is $\omega=2\pi/T$, so its centripetal acceleration is $\omega^2r=(2\pi/T)^2r$. Equating the gravitational force to the required centripetal force: $\frac{GMm}{r^2} = m\left(\frac{2\pi}{T}\right)^2r$

Cancelling $m$ from both sides: $\frac{GM}{r^2} = \frac{4\pi^2}{T^2}r$

Multiplying both sides by $r^2$: $GM = \frac{4\pi^2r^3}{T^2}$

Rearranging for $r^3$: $r^3 = \frac{GMT^2}{4\pi^2}$

Taking the cube root of both sides: $r = \left(\frac{GMT^2}{4\pi^2}\right)^{1/3}$

as required.

Part (c): Radius of the geostationary orbit

First convert the rotation period to seconds: $T = 30.0\text{ hours} = 30.0\times3600 = 1.08\times10^5\text{ s}$

With $M=5.40\times10^{24}\text{ kg}$: $GM = (6.67\times10^{-11})(5.40\times10^{24})$

$6.67\times5.40=36.018$, and $10^{-11}\times10^{24}=10^{13}$, so $GM=3.6018\times10^{14}\text{ m}^3\text{ s}^{-2}$.

$T^2 = (1.08\times10^5)^2 = 1.1664\times10^{10}\text{ s}^2$

$GMT^2 = 3.6018\times10^{14}\times1.1664\times10^{10} = 4.2011\times10^{24}$

(Working the coefficient: $3.6018\times1.1664=4.2011$; working the powers: $10^{14}\times10^{10}=10^{24}$.)

Dividing by $4\pi^2=39.478$: $r^3 = \frac{4.2011\times10^{24}}{39.478} = 1.0642\times10^{23}\text{ m}^3$

Taking the cube root. Writing $1.0642\times10^{23}=106.42\times10^{21}$, so that the power of ten is a multiple of $3$: $r = \sqrt[3]{106.42}\times10^{7} = 4.739\times10^{7}\text{ m} \approx 4.74\times10^7\text{ m}$

Check by cubing the answer back: $(4.739\times10^7)^3 = 4.739^3\times10^{21}=106.4\times10^{21}=1.064\times10^{23}\text{ m}^3$, which matches $r^3$ found above. Also, as a sanity check, Halvane’s mass is somewhat less than Earth’s ($5.97\times10^{24}\text{ kg}$) but its rotation period ($30.0$ hours) is longer than Earth’s ($24$ hours), and Earth’s real geostationary radius is about $4.22\times10^7\text{ m}$, so a somewhat larger radius here ($4.74\times10^7\text{ m}$) is physically reasonable.

Part (d): Orbital speed in the geostationary orbit

Using $v=2\pi r/T$: $v = \frac{2\pi(4.739\times10^7)}{1.08\times10^5} = \frac{2.9776\times10^8}{1.08\times10^5} = 2757\text{ m s}^{-1} \approx 2.76\times10^3\text{ m s}^{-1}$

Check using $v=\sqrt{GM/r}$ instead: $GM/r=3.6018\times10^{14}/(4.739\times10^7)=7.600\times10^6$; $\sqrt{7.600\times10^6}=2757\text{ m s}^{-1}$. Both methods agree.

Final answers

• (a) Orbit in the plane of Halvane’s equator; orbit in the same direction as Halvane’s rotation (west to east)
• (b) $r=\boxed{(GMT^2/4\pi^2)^{1/3}}$ (derived from equating gravitational and centripetal force, with $v=2\pi r/T$)
• (c) $r = \boxed{4.74\times10^7}\text{ m}$
• (d) $v = \boxed{2.76\times10^3}\text{ m s}^{-1}$