Escape Velocity: notes and previous year questions

Answer: Option D

$v_{e} = \sqrt{\dfrac{2GM}{R}}$, and the masses are equal, so only the radii matter:

$\dfrac{v_{2}}{v_{1}} = \sqrt{\dfrac{R_{1}}{R_{2}}} = \sqrt{2}$

which is D. The same mass squeezed into a smaller planet is harder to leave.

The trap is A, $2$, which forgets the square root. B, $\dfrac{1}{\sqrt{2}}$, turns the ratio upside down, as if a smaller planet were easier to leave. C, $1$, assumes equal masses mean equal escape velocities; the radius matters too.

Answer: Option D

$v_{e} = R\sqrt{\dfrac{8\pi G\rho}{3}}$, so $v_{e} \propto R\sqrt{\rho}$. For planet $B$:

$\dfrac{v_{B}}{v_{A}} = 0.1 \times \sqrt{0.1} = \dfrac{1}{10\sqrt{10}}$

With $v_{A} = 10\ \mathrm{km/s} = 10\,000\ \mathrm{m/s}$:

$v_{B} = \dfrac{10\,000}{10\sqrt{10}} = \dfrac{1000}{\sqrt{10}} = 100\sqrt{10} \approx 316\ \mathrm{m/s}$

which is D. The answer is asked in $\mathrm{m/s}$, so convert the $10\ \mathrm{km/s}$ first.

The trap is A, $1000$: it scales with the radius alone and forgets the density. B and C, $200\sqrt{5} \approx 447$ and $1000\sqrt{2} \approx 1414$, do not follow from $R\sqrt{\rho}$ at all.

Answer: Options B, D

Each fixed body is a distance $L$ from the midpoint, so the potential there is $-\dfrac{GM}{L} - \dfrac{GM}{L} = -\dfrac{2GM}{L}$. To escape from both,

$\dfrac{1}{2}v^{2} = \dfrac{2GM}{L}$, so $v = 2\sqrt{\dfrac{GM}{L}}$

which is B. Firing it perpendicular to the line changes nothing: energy has no direction. And since the two bodies are fixed and gravity is conservative, the mechanical energy of $m$, kinetic plus potential, stays the same as it flies off: D.

The trap is C, $\sqrt{\dfrac{2GM}{L}}$, which counts only one of the two bodies. A, $4\sqrt{\dfrac{GM}{L}}$, would need $\dfrac{1}{2}v^{2} = \dfrac{8GM}{L}$, four times the potential energy there is to pay back.

Answer: Options A, C, D

A: $2gR = 2 \times 9.8 \times 6.4 \times 10^{6} = 1.2544 \times 10^{8}\ \mathrm{m^{2}/s^{2}}$, and its square root is $1.12 \times 10^{4}\ \mathrm{m/s}$.

C: that is exactly the speed of arrival from rest at a great distance, $\sqrt{\dfrac{2GM}{R}}$.

D: in $h = \dfrac{Rv^{2}}{2gR - v^{2}}$, doubling $v$ makes the top four times larger, and the bottom smaller as well: $2gR - v^{2}$ falls from $125.4 - 9 = 116.4$ to $125.4 - 36 = 89.4$ (in $\mathrm{km^{2}/s^{2}}$). The height grows about $5.2$ times.

B is the trap: $m$ cancels from the energy equation, so the height depends only on the launch speed, not on the mass. A heavier body needs more energy, but not more speed.

Answer: Option A

The probe is $r = R + 8R = 9R$ from the centre, so

$v = \sqrt{\dfrac{2GM}{9R}} = \dfrac{1}{3}\sqrt{\dfrac{2GM}{R}} = \dfrac{v_{e}}{3}$

which is A.

The trap is C, $\dfrac{v_{e}}{\sqrt{8}}$, which uses the height $8R$ as the distance from the centre. B, $\dfrac{v_{e}}{9}$, forgets the square root, and D makes both mistakes at once.

Answer: 2.3 km

$\dfrac{8\pi G\rho}{3} = \dfrac{8 \times 3.14 \times 6.67 \times 10^{-11} \times 3000}{3} \approx 1.68 \times 10^{-6}\ \mathrm{s^{-2}}$, whose square root is about $1.29 \times 10^{-3}\ \mathrm{s^{-1}}$. Setting $v_{e} = 3\ \mathrm{m/s}$:

$R = \dfrac{3}{1.29 \times 10^{-3}} \approx 2.3 \times 10^{3}\ \mathrm{m} = 2.3\ \mathrm{km}$

Any bigger, and the jumper comes back down. A good jump on a rock a few kilometres across is a trip to space.

The trap is forgetting the square root, which gives $\dfrac{3}{1.68 \times 10^{-6}}\ \mathrm{m}$, nearly two thousand kilometres.