Space Physics: Question 10

Worked solution

Part (a): Orbital period of Wayfarer-3

The orbital speed formula is:

$v = \frac{2\pi r}{T}$

Rearranging for the orbital period, $T$:

$T = \frac{2\pi r}{v}$

Substituting the values for Wayfarer-3:

$T = \frac{2\pi \times 4.00\times10^{7}\text{ m}}{5.03\times10^{2}\text{ m/s}}$

$T = \frac{2.513\times10^{8}\text{ m}}{5.03\times10^{2}\text{ m/s}}$

$T = \boxed{5.00\times10^{5}\text{ s}}\ \text{(3 s.f.)}$

Converting to hours (dividing by 3600, since there are 3600 seconds in an hour):

$T = \frac{5.00\times10^{5}\text{ s}}{3600\text{ s/h}} = \boxed{139\text{ hours}}\ \text{(3 s.f.)}$

Part (b): Why Ithel’s orbital period is much shorter

Ithel orbits at a much smaller average radius than Wayfarer-3. Two effects combine to make its orbital period much shorter:

• Distance travelled: the distance covered in one orbit is the circumference of the orbit, $2\pi r$. Since Ithel’s orbital radius $r$ is much smaller, the circumference it must travel in one orbit is also much smaller than Wayfarer-3’s.
• Orbital speed: Ithel orbits much closer to Ostrena, where the strength of Ostrena’s gravitational field is greater. A stronger gravitational field requires a greater orbital speed to maintain a stable orbit, so Ithel’s average orbital speed is greater than Wayfarer-3’s.

Since the orbital period is $T = \dfrac{2\pi r}{v}$, Ithel has both a smaller distance to travel ($2\pi r$) and a greater speed ($v$) at which to travel it. Both effects reduce $T$, so Ithel’s orbital period is much shorter than Wayfarer-3’s.

Final answers

• (a) Orbital period of Wayfarer-3 $=$ $5.00\times10^5\text{ s}$ (3 s.f.) $=$ 139 hours (3 s.f.)
• (b) Ithel’s period is much shorter because it travels a smaller distance ($2\pi r$) in one orbit and does so at a greater orbital speed, both of which reduce $T = \dfrac{2\pi r}{v}$