Gravitational Fields: Question 10

Worked solution

Part (a): Orbital speed of Probe 1

For a circular orbit, $v=\sqrt{GM/r}$. With $M=6.40\times10^{23}\text{ kg}$ and $r_1=1.00\times10^7\text{ m}$:

$GM = (6.67\times10^{-11})(6.40\times10^{23})$

$6.67\times6.40=42.688$, and $10^{-11}\times10^{23}=10^{12}$, so $GM=4.2688\times10^{13}\text{ m}^3\text{ s}^{-2}$.

$\frac{GM}{r_1} = \frac{4.2688\times10^{13}}{1.00\times10^7} = 4.2688\times10^6$

$v_1 = \sqrt{4.2688\times10^6} = 2066\text{ m s}^{-1} \approx 2.07\times10^3\text{ m s}^{-1}$

Check: $\sqrt{4.2688}\approx2.0661$ and $\sqrt{10^6}=10^3$, so $v_1\approx2.0661\times10^3=2066\text{ m s}^{-1}$, matching.

Part (b): Orbital period of Probe 1

Using $T=2\pi r_1/v_1$: $T_1 = \frac{2\pi(1.00\times10^7)}{2066} = \frac{6.2832\times10^7}{2066} = 3.0412\times10^4\text{ s}$

Converting to hours (dividing by $3600\text{ s}$ per hour): $T_1 = \frac{3.0412\times10^4}{3600} = 8.448\text{ hours} \approx 8.45\text{ hours}$

Check using $T=2\pi\sqrt{r^3/GM}$ directly: $r_1^3=(1.00\times10^7)^3=1.00\times10^{21}\text{ m}^3$; $r_1^3/GM=1.00\times10^{21}/4.2688\times10^{13}=2.3426\times10^7$; $\sqrt{2.3426\times10^7}=4840$; $T_1=2\pi\times4840=3.041\times10^4\text{ s}=8.45$ hours. Both methods agree.

Part (c): Orbital period of Probe 2 via Kepler’s third law

Both probes orbit the same planet, so $T^2\propto r^3$ applies with the same constant of proportionality for both: $\left(\frac{T_2}{T_1}\right)^2 = \left(\frac{r_2}{r_1}\right)^3$

Since $r_2=4.00\times10^7\text{ m}=4r_1$: $\left(\frac{T_2}{T_1}\right)^2 = 4^3 = 64 \quad\Rightarrow\quad \frac{T_2}{T_1} = \sqrt{64} = 8$

$T_2 = 8\times T_1 = 8\times8.448 = 67.58\text{ hours} \approx 67.6\text{ hours}$

Check by recalculating directly: $r_2^3=(4.00\times10^7)^3=6.40\times10^{22}\text{ m}^3$; $r_2^3/GM=6.40\times10^{22}/4.2688\times10^{13}=1.4991\times10^9$; $\sqrt{1.4991\times10^9}=3.8718\times10^4$; $T_2=2\pi\times3.8718\times10^4=2.4327\times10^5\text{ s}=67.58$ hours. This matches the ratio method, confirming $T_2\approx67.6$ hours.

Part (d): Comparing the orbital speeds

Since $v=\sqrt{GM/r}$, orbital speed is inversely proportional to the square root of the orbital radius: $v\propto1/\sqrt{r}$. As $r_2=4r_1$: $v_2 = \frac{v_1}{\sqrt{4}} = \frac{v_1}{2}$

So Probe 2, in the larger orbit, moves at half the orbital speed of Probe 1.

Check by calculating $v_2$ directly: $GM/r_2=4.2688\times10^{13}/(4.00\times10^7)=1.0672\times10^6$; $v_2=\sqrt{1.0672\times10^6}=1033\text{ m s}^{-1}$. Comparing with $v_1=2066\text{ m s}^{-1}$: $v_2/v_1=1033/2066=0.500$, confirming $v_2=v_1/2$.

Final answers

• (a) $v_1 = \boxed{2.07\times10^3}\text{ m s}^{-1}$
• (b) $T_1 = \boxed{8.45}$ hours
• (c) $T_2 = \boxed{67.6}$ hours
• (d) Probe 2 orbits at half the speed of Probe 1 ($v_2=v_1/2$), since $v\propto1/\sqrt{r}$ and $r_2=4r_1$