Gravitational Fields: Question 4

Worked solution

Part (a): Deriving the orbital speed

For the moon (mass $m$) to move in a circular orbit of radius $r$ at constant speed $v$, the gravitational force from Draymoor must provide exactly the centripetal force needed: $\frac{GMm}{r^2} = \frac{mv^2}{r}$

Cancelling the moon’s mass $m$ from both sides (it appears on both sides, so the orbit does not depend on the moon’s own mass): $\frac{GM}{r^2} = \frac{v^2}{r}$

Multiplying both sides by $r$: $\frac{GM}{r} = v^2$

Taking the square root of both sides: $v = \sqrt{\frac{GM}{r}}$

as required.

Part (b): Orbital speed of the moon

With $M=8.10\times10^{24}\text{ kg}$ and $r=5.00\times10^8\text{ m}$: $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{(6.67\times10^{-11})(8.10\times10^{24})}{5.00\times10^8}}$

Numerator: $6.67\times8.10=54.027$, and $10^{-11}\times10^{24}=10^{13}$, so $GM=5.4027\times10^{14}$.

$\frac{GM}{r} = \frac{5.4027\times10^{14}}{5.00\times10^8} = 1.08054\times10^6$

$v = \sqrt{1.08054\times10^6} = 1039.5\text{ m s}^{-1} \approx 1.04\times10^3\text{ m s}^{-1}$

Check by recomputing differently: $\sqrt{1.08054}\approx1.0395$ and $\sqrt{10^6}=10^3$, so $v\approx1.0395\times10^3=1039.5\text{ m s}^{-1}$, matching exactly.

Part (c): Orbital period

Using $T=2\pi r/v$ (the circumference of the orbit divided by the speed): $T = \frac{2\pi r}{v} = \frac{2\pi(5.00\times10^8)}{1039.5} = \frac{3.1416\times10^9}{1039.5} = 3.0224\times10^6\text{ s}$

Converting to days (dividing by $86400\text{ s}$ per day): $T = \frac{3.0224\times10^6}{86400} = 34.98\text{ days} \approx 35.0\text{ days}$

Check using $T=2\pi\sqrt{r^3/(GM)}$ directly: $r^3=(5.00\times10^8)^3=1.25\times10^{26}\text{ m}^3$; $r^3/GM=1.25\times10^{26}/5.4027\times10^{14}=2.3138\times10^{11}$; $\sqrt{2.3138\times10^{11}}=4.810\times10^5$; $T=2\pi\times4.810\times10^5=3.022\times10^6\text{ s}=34.98$ days. Both methods agree.

Part (d): Radius of the second moon’s orbit

Both moons orbit the same planet, Draymoor, so Kepler’s third law applies to both in the form $T^2=\dfrac{4\pi^2}{GM}r^3$, i.e. $T^2\propto r^3$ with the same constant of proportionality. For the two moons: $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$

Since $T_2=2T_1$: $2^2 = \left(\frac{r_2}{r_1}\right)^3 \quad\Rightarrow\quad \frac{r_2}{r_1} = 4^{1/3} = 1.587$

$r_2 = 1.587\times(5.00\times10^8) = 7.937\times10^8\text{ m} \approx 7.94\times10^8\text{ m}$

Check by verifying $T_2$ directly at this radius: $r_2^3=(7.937\times10^8)^3=5.001\times10^{26}\text{ m}^3$; $r_2^3/GM=5.001\times10^{26}/5.4027\times10^{14}=9.258\times10^{11}$; $T_2=2\pi\sqrt{9.258\times10^{11}}=2\pi\times9.622\times10^5=6.046\times10^6\text{ s}$. Comparing with $2T=2\times3.022\times10^6=6.044\times10^6\text{ s}$, the two values agree (to rounding), confirming $r_2\approx7.94\times10^8\text{ m}$.

Final answers

• (a) $v=\boxed{\sqrt{GM/r}}$ (derived by equating gravitational and centripetal force)
• (b) $v = \boxed{1.04\times10^3}\text{ m s}^{-1}$
• (c) $T = \boxed{35.0}$ days
• (d) $r_2 = \boxed{7.94\times10^8}\text{ m}$