How to Calculate Escape Speed from a Planet’s Mass and Radius

A planet has a mass of \(6.0 \times 10^{24}\,\text{kg}\) and a radius of \(6.4 \times 10^6\,\text{m}\). How fast must a rocket leave its surface to escape the planet's gravitational field?

You don't need to know anything about the rocket's mass, its engine design, or how long it accelerates. If you know the planet's mass and radius, you can calculate the minimum escape speed directly:

\[
\boxed{v_{\text{esc}}=\sqrt{\frac{2GM}{R}}}
\]

Here:
- \(v_{\text{esc}}\) is escape speed in metres per second (\(\text{m s}^{-1}\)).
- \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}\).
- \(M\) is the mass of the planet in kilograms (\(\text{kg}\)).
- \(R\) is the distance from the planet's centre to the launch point in metres (\(\text{m}\)).

For a launch from the planet's surface, \(R\) is simply the planet's radius.

The important part isn't memorising the equation. It's understanding why the planet's mass increases escape speed, why its radius decreases it, and why gravitational potential energy must be treated as negative.

## How to calculate escape speed

You can solve almost any standard HSC escape-speed calculation using three steps.

**Step 1:** Identify the planet's mass, \(M\), and radius, \(R\).

**Step 2:** Convert both quantities into SI units: kilograms and metres.

**Step 3:** Substitute into:

\[
v_{\text{esc}}=\sqrt{\frac{2GM}{R}}
\]

Calculate the value inside the square root before taking the square root. Your final answer should be in \(\text{m s}^{-1}\).

[IMAGE: An annotated scientific illustration of a small rocket launching vertically from the surface of a spherical planet. Show the planet's centre, its radius R extending to the launch point, the planet's mass M, and the rocket's outward velocity labelled v. Include a subtle indication that gravitational attraction acts towards the planet's centre. Keep the geometry clear and distinguish the launch velocity from the inward gravitational force.]

### Worked example: Calculate the escape speed from a planet

A fictional planet has a mass of \(6.0\times10^{24}\,\text{kg}\) and a radius of \(6.4\times10^6\,\text{m}\). Calculate the minimum speed required for an object launched from its surface to escape its gravitational field.

**Step 1**: Identify the given quantities.

\[
\begin{aligned}
M &= 6.0\times10^{24}\,\text{kg}\\
R &= 6.4\times10^6\,\text{m}\\
G &= 6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}
\end{aligned}
\]

**Step 2**: Write the escape-speed equation.

\[
v_{\text{esc}}=\sqrt{\frac{2GM}{R}}
\]

**Step 3**: Substitute the values.

\[
v_{\text{esc}}=
\sqrt{
\frac{
2(6.67\times10^{-11})(6.0\times10^{24})
}{
6.4\times10^6
}
}
\]

**Step 4**: Calculate.

\[
\begin{aligned}
v_{\text{esc}}&=\sqrt{1.2506\times10^8}\\
&=1.118\times10^4\,\text{m s}^{-1}
\end{aligned}
\]

Therefore:

\[
\boxed{v_{\text{esc}}=1.12\times10^4\,\text{m s}^{-1}}
\]

The object needs an initial speed of approximately \(11.2\,\text{km s}^{-1}\) to escape.

This is a minimum. At exactly this speed, the object continues moving outwards, slowing down as it climbs, and approaches zero speed only as its distance from the planet approaches infinity.

## Why the formula works: gravitational potential energy

There is an important detail hidden inside the escape-speed equation: gravity does not suddenly stop acting when an object gets far away.

Imagine launching a probe vertically from a planet. As it moves away:
- Its gravitational potential energy increases.
- Its kinetic energy decreases.
- Its total mechanical energy remains constant, provided gravity is the only force doing work.

To understand why, we need to choose a reference point for gravitational potential energy.

### Why is gravitational potential energy negative?

We define gravitational potential energy to be zero at infinity. This is a convenient reference point because gravitational attraction approaches zero as the separation between two objects becomes infinitely large.

For an object of mass \(m\), at a distance \(r\) from the centre of a planet of mass \(M\):

\[
U=-\frac{GMm}{r}
\]

Here:
- \(U\) is gravitational potential energy in joules (J).
- \(m\) is the mass of the escaping object in kilograms.
- \(r\) is its distance from the planet's centre in metres.

The negative sign tells us that the object has less gravitational potential energy than it would have infinitely far away.

As the object moves outwards, its potential energy becomes less negative and approaches zero.

[IMAGE: A scientific illustration of a spherical planet with a probe at its surface and a second probe position much farther away. Show the centre-to-object distance r, the surface radius R, and an arrow indicating outward motion. Include an infinity reference at the far end of the scene, with gravitational potential energy approaching zero. Clearly distinguish zero potential energy at infinity from the non-zero gravitational attraction at finite distances.]

### How does this give us escape speed?

The total mechanical energy of the object is the sum of its kinetic and gravitational potential energies:

\[
E=K+U
\]

Therefore:

\[
E=\frac12 mv^2-\frac{GMm}{r}
\]

For an object launched from the planet's surface, \(r=R\).

Now consider the minimum speed needed to escape.

At exactly escape speed, the object reaches infinity with its speed approaching zero. Both its kinetic energy and gravitational potential energy approach zero.

Therefore, its total mechanical energy must equal zero:

\[
\frac12 mv_{\text{esc}}^2-\frac{GMm}{R}=0
\]

Rearranging:

\[
\frac12 mv_{\text{esc}}^2=\frac{GMm}{R}
\]

The escaping object's mass, \(m\), appears on both sides, so it cancels:

\[
v_{\text{esc}}^2=\frac{2GM}{R}
\]

Taking the square root gives:

\[
\boxed{v_{\text{esc}}=\sqrt{\frac{2GM}{R}}}
\]

This is why escape speed depends on the planet's mass and radius, but not on the mass of the escaping object.

The full derivation and its connection to orbital speed are covered in [Escape Velocity for HSC Physics: Derivation and Calculations](https://tutorgum.com/learn/resources/hsc-physics-escape-velocity/).

### Bound energy is not the same as kinetic energy

This is a common source of mistakes in HSC questions.

Kinetic energy cannot be negative:

\[
K=\frac12 mv^2\geq 0
\]

However, total mechanical energy can be negative because gravitational potential energy is negative.

An object can therefore have positive kinetic energy while still being gravitationally bound.

For an object moving under gravity alone:

| Total mechanical energy | What happens? |
|---|---|
| \(E<0\) | The object is gravitationally bound. |
| \(E=0\) | The object is at the minimum escape-energy threshold. |
| \(E>0\) | The object can escape with kinetic energy remaining at infinity. |

A useful way to remember this is that kinetic energy describes motion, while total mechanical energy determines whether the object has enough energy to escape.

[IMAGE: An educational comparison of three probe trajectories around the same spherical planet. Show one probe with insufficient launch energy reaching a finite turning point, one at the escape threshold continuing towards infinity, and one with excess energy continuing outwards. Include compact labels E less than zero, E equals zero, and E greater than zero. Keep the trajectories physically distinct and avoid presenting the illustration as a measured result.]

For a more detailed treatment of how kinetic energy, gravitational potential energy, and total energy change with distance, see [How to Read Gravitational Energy Graphs for Escape Speed](https://tutorgum.com/learn/resources/how-to-read-gravitational-energy-graphs-for-escape-from-a-planet/).

## A common calculation trap: radius versus height

The \(R\) in the escape-speed equation represents the distance from the planet's centre to the launch point.

It does not always mean the planet's surface radius.

If an object launches from a point above the surface, you must add its height to the planet's radius:

\[
r=R+h
\]

where \(h\) is the height above the surface.

This matters because gravitational attraction depends on distance from the centre of the planet.

### Worked example: Escape speed from a high-altitude launch point

A fictional planet has a mass of \(8.0\times10^{24}\,\text{kg}\) and a radius of \(4.0\times10^6\,\text{m}\).

A probe launches from a platform \(1.0\times10^6\,\text{m}\) above the planet's surface.

Calculate:

a) The escape speed from the surface.

b) The escape speed from the elevated platform.

**Step 1**: Identify the given information.

\[
\begin{aligned}
M&=8.0\times10^{24}\,\text{kg}\\
R&=4.0\times10^6\,\text{m}\\
h&=1.0\times10^6\,\text{m}
\end{aligned}
\]

**Step 2**: Calculate the surface escape speed.

At the surface, the distance from the centre is simply \(R\).

\[
v_{\text{esc}}=\sqrt{\frac{2GM}{R}}
\]

Substituting:

\[
v_{\text{esc}}=
\sqrt{
\frac{
2(6.67\times10^{-11})(8.0\times10^{24})
}{
4.0\times10^6
}
}
\]

\[
v_{\text{esc}}=\sqrt{2.668\times10^8}
\]

\[
\boxed{v_{\text{esc}}=1.63\times10^4\,\text{m s}^{-1}}
\]

**Step 3**: Calculate the distance from the centre at the platform.

The platform is above the surface, so:

\[
r=R+h
\]

\[
r=4.0\times10^6+1.0\times10^6
\]

\[
r=5.0\times10^6\,\text{m}
\]

**Step 4**: Calculate the escape speed from the platform.

Use the same equation, replacing \(R\) with \(r\):

\[
v_{\text{esc}}=\sqrt{\frac{2GM}{r}}
\]

\[
v_{\text{esc}}=
\sqrt{
\frac{
2(6.67\times10^{-11})(8.0\times10^{24})
}{
5.0\times10^6
}
}
\]

\[
\boxed{v_{\text{esc}}=1.46\times10^4\,\text{m s}^{-1}}
\]

The escape speed from the elevated platform is approximately \(14.6\,\text{km s}^{-1}\), compared with \(16.3\,\text{km s}^{-1}\) at the surface.

The higher launch point requires less initial speed because the probe is already farther from the planet's centre.

Notice that we used \(5.0\times10^6\,\text{m}\), not the platform height of \(1.0\times10^6\,\text{m}\). Using height alone would give an incorrect result.

## What happens when the planet's mass or radius changes?

The equation also gives us a quick way to compare different planets without calculating their escape speeds from scratch.

\[
v_{\text{esc}}=\sqrt{\frac{2GM}{R}}
\]

Since \(G\) and 2 are constants:

\[
v_{\text{esc}}\propto\sqrt{\frac{M}{R}}
\]

This means:
- Increasing planetary mass increases escape speed.
- Increasing planetary radius decreases escape speed if mass stays constant.
- Increasing both mass and radius by the same factor leaves escape speed unchanged.

The square root is important. Escape speed does not increase directly in proportion to mass or decrease directly in proportion to radius.

For two planets, the most useful comparison equation is:

\[
\boxed{
\frac{v_2}{v_1}
=
\sqrt{
\frac{M_2}{M_1}
\frac{R_1}{R_2}
}
}
\]

You can use this to compare planets even when their masses and radii are given as multiples rather than actual numerical values.

For example:

| Change | Effect on escape speed |
|---|---|
| Mass increases by a factor of 4, radius unchanged | Doubles |
| Radius increases by a factor of 4, mass unchanged | Halves |
| Mass and radius both increase by a factor of 4 | Unchanged |

[IMAGE: Two spherical planets shown side by side, one smaller and one larger, with their centres, radii and relative masses clearly distinguished. Include a small comparison of how increasing mass strengthens gravitational binding while increasing radius changes the starting distance. Avoid suggesting that size alone determines escape speed.]

### A note on equal-density planets

There is an interesting consequence when two planets have the same average density.

The mass of a spherical planet is:

\[
M=\rho\frac43\pi R^3
\]

Here, \(\rho\) is the planet's average density in \(\text{kg m}^{-3}\).

Substituting into the escape-speed equation:

\[
v_{\text{esc}}
=
\sqrt{
\frac{2G}{R}
\left(\rho\frac43\pi R^3\right)
}
\]

Simplifying:

\[
v_{\text{esc}}=\sqrt{\frac{8\pi G\rho}{3}R^2}
\]

Therefore, for planets with equal average density:

\[
\boxed{v_{\text{esc}}\propto R}
\]

This is a useful result to remember. A larger planet does not necessarily have a lower escape speed. The effect depends on how its mass changes with its radius.

## Where does the gravitational constant come from?

The value of \(G\) in the escape-speed equation is not an arbitrary number. It represents the strength of gravitational attraction between masses.

Its experimental history is connected to Henry Cavendish's torsion-balance experiment, published in 1798.

The original problem was difficult: how could scientists measure the extremely weak gravitational attraction between ordinary objects in a laboratory?

[IMAGE: A historically informed scientific illustration of the Cavendish torsion-balance apparatus. Show a horizontal rod suspended from a thin vertical fibre, with two small lead spheres attached at opposite ends. Position two much larger lead source spheres nearby on a separate support. Include a small mirror attached to the suspension and a reflected light beam travelling towards a distant measuring scale. Show the physical arrangement clearly, including the movement of the large spheres between alternative positions.]

The apparatus worked as follows.

**Step 1**: Two small lead spheres were attached to opposite ends of a horizontal rod. The rod was suspended from a thin fibre, allowing it to rotate when the fibre twisted.

**Step 2**: Two much larger lead spheres were positioned close to the smaller spheres. Their gravitational attraction produced a tiny turning effect on the suspended rod.

**Step 3**: The fibre twisted until its restoring torque balanced the gravitational torque. A mirror and optical measuring arrangement made the small angular movement easier to observe.

**Step 4**: The large spheres were moved to alternative positions. The predicted change in gravitational attraction was a reversal of the twisting effect. The measured deflection provided evidence for the gravitational attraction between ordinary masses.

The main experimental difficulty was the extremely small signal. Temperature differences, magnetic effects, and other disturbances could interfere with the measurement. Careful shielding, repositioning the masses, and checking possible sources of error helped improve the reliability of the result.

Cavendish's original aim was to determine the average density of Earth. His measurements are now recognised as a landmark in determining the universal gravitational constant. You can read more in the [American Physical Society's historical account](https://www.aps.org/apsnews/2008/06/1798-cavendish-weighs-world).

The important distinction is that this experiment did not directly measure escape speed. It helped establish the gravitational constant that allows us to calculate gravitational effects, including the minimum escape speed from a planet.

## Assumptions behind the escape-speed calculation

The equation is based on an idealised gravitational model.

For standard HSC calculations, assume:

- The planet is spherical and has a uniform external gravitational field consistent with its total mass.
- The planet's mass remains constant.
- The escaping object has negligible mass compared with the planet.
- Air resistance is negligible.
- Gravitational effects from other celestial bodies are ignored.
- No additional propulsion acts after the object receives its initial speed.
- Mechanical energy is conserved.

In a real rocket launch, the situation is more complicated. Atmospheric drag, planetary rotation, engine thrust, and changing rocket mass all affect the launch.

The escape-speed equation gives the ideal minimum initial speed at a particular distance. It is not a complete calculation of the fuel or energy needed for an actual space mission.

## Questions and solutions

All planet masses and radii in the following questions are illustrative values. Use \(G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}\) unless otherwise stated.

### Question 1

A fictional planet has a mass of \(4.0\times10^{24}\,\text{kg}\) and a radius of \(5.0\times10^6\,\text{m}\).

Calculate the minimum escape speed from the planet's surface.

### Solution 1

The minimum escape speed is \(1.03\times10^4\,\text{m s}^{-1}\), or approximately \(10.3\,\text{km s}^{-1}\).

Use:

\[
v_{\text{esc}}=\sqrt{\frac{2GM}{R}}
\]

Substitute the given values:

\[
v_{\text{esc}}=
\sqrt{
\frac{
2(6.67\times10^{-11})(4.0\times10^{24})
}{
5.0\times10^6
}
}
\]

Simplifying:

\[
v_{\text{esc}}=\sqrt{1.0672\times10^8}
\]

\[
\boxed{v_{\text{esc}}=1.03\times10^4\,\text{m s}^{-1}}
\]

The result is in metres per second, as expected from the SI units in the equation.

### Question 2

Two fictional planets, P and Q, have the following properties:

| Property | Planet P | Planet Q |
|---|---:|---:|
| Mass | \(4.0\times10^{24}\,\text{kg}\) | \(9.0\times10^{24}\,\text{kg}\) |
| Radius | \(3.0\times10^6\,\text{m}\) | \(9.0\times10^6\,\text{m}\) |

A student claims that Planet Q must have the greater escape speed because it has more than twice the mass of Planet P.

Determine which planet actually has the greater escape speed and calculate the ratio of their escape speeds.

[IMAGE: Side-by-side comparison of fictional planets P and Q. Show Planet Q with three times the radius and 2.25 times the mass of Planet P. Label the two radii and masses using the values supplied in the question. Do not show the escape speeds or indicate which planet wins the comparison.]

### Solution 2

Planet P has the greater escape speed. Its escape speed is approximately 1.15 times that of Planet Q.

The student's prediction considers mass but ignores radius. We need to account for both quantities.

**Step 1**: Write the ratio equation.

\[
\frac{v_Q}{v_P}
=
\sqrt{
\frac{M_Q}{M_P}
\frac{R_P}{R_Q}
}
\]

**Step 2**: Substitute the mass and radius ratios.

\[
\frac{v_Q}{v_P}
=
\sqrt{
\frac{9.0}{4.0}
\times
\frac{3.0}{9.0}
}
\]

\[
=
\sqrt{0.75}
\]

\[
\frac{v_Q}{v_P}=0.866
\]

**Step 3**: Interpret the result.

Planet Q's escape speed is approximately 86.6% of Planet P's escape speed.

Therefore:

\[
\frac{v_P}{v_Q}
=
\frac{1}{0.866}
\]

\[
\boxed{\frac{v_P}{v_Q}=1.15}
\]

Planet Q has greater mass, but its radius increases by an even greater factor relative to its mass. Its mass-to-radius ratio is therefore smaller.

This is why comparing planetary mass alone is not enough.

### Question 3

A fictional planet has a mass of \(5.0\times10^{24}\,\text{kg}\) and a radius of \(5.0\times10^6\,\text{m}\).

A probe of mass \(120\,\text{kg}\) is launched vertically outwards from the planet's surface at \(8.0\,\text{km s}^{-1}\).

Assume that the probe experiences only the planet's gravitational attraction after launch.

Calculate:

a) The planet's surface escape speed.

b) The probe's total mechanical energy immediately after launch.

c) Whether the probe escapes or remains gravitationally bound.

d) The maximum distance the probe reaches from the planet's centre.

[IMAGE: A spherical planet with a probe launched vertically outwards from its surface. Label the planet's radius R, mass M, the probe's mass 120 kg, and its initial outward speed of 8.0 km/s. Show the radial path extending away from the planet and a possible turning-point location marked as an unknown r_max. Do not reveal the calculated turning point.]

### Solution 3

The probe does not escape. It remains gravitationally bound and reaches a maximum distance of approximately \(9.61\times10^6\,\text{m}\) from the planet's centre.

**Step 1**: Calculate the surface escape speed.

\[
v_{\text{esc}}=\sqrt{\frac{2GM}{R}}
\]

Substituting:

\[
v_{\text{esc}}
=
\sqrt{
\frac{
2(6.67\times10^{-11})(5.0\times10^{24})
}{
5.0\times10^6
}
}
\]

\[
\boxed{v_{\text{esc}}=1.15\times10^4\,\text{m s}^{-1}}
\]

The probe's initial speed is \(8.0\times10^3\,\text{m s}^{-1}\), which is below escape speed.

**Step 2**: Calculate the initial kinetic energy.

\[
K=\frac12 mv^2
\]

\[
K=\frac12(120)(8.0\times10^3)^2
\]

\[
\boxed{K=3.84\times10^9\,\text{J}}
\]

**Step 3**: Calculate the initial gravitational potential energy.

Remember that the zero of gravitational potential energy is at infinity.

\[
U=-\frac{GMm}{R}
\]

\[
U=
-\frac{
(6.67\times10^{-11})(5.0\times10^{24})(120)
}{
5.0\times10^6
}
\]

\[
\boxed{U=-8.00\times10^9\,\text{J}}
\]

**Step 4**: Calculate total mechanical energy.

\[
E=K+U
\]

\[
E=3.84\times10^9-8.00\times10^9
\]

\[
\boxed{E=-4.16\times10^9\,\text{J}}
\]

The negative total mechanical energy confirms that the probe is gravitationally bound.

Notice that the probe still has positive kinetic energy. It is moving upwards, but it does not have enough total energy to escape.

**Step 5**: Find the maximum distance from the planet's centre.

At the maximum distance, the probe momentarily stops before moving back towards the planet.

Therefore:

\[
K=0
\]

Its total mechanical energy is now entirely gravitational potential energy:

\[
E=-\frac{GMm}{r_{\max}}
\]

Rearrange:

\[
r_{\max}=-\frac{GMm}{E}
\]

Substitute:

\[
r_{\max}
=
-\frac{
(6.67\times10^{-11})(5.0\times10^{24})(120)
}{
-4.16\times10^9
}
\]

\[
\boxed{r_{\max}\approx9.62\times10^6\,\text{m}}
\]

This is the maximum distance from the centre, not the height above the surface.

The maximum height above the surface is approximately:

\[
h_{\max}=r_{\max}-R
\]

\[
h_{\max}=9.62\times10^6-5.0\times10^6
\]

\[
\boxed{h_{\max}=4.62\times10^6\,\text{m}}
\]

The probe rises to this height, momentarily stops, and then falls back.

### Question 4

Two spherical planets, A and B, have the same average density.

Planet A has a radius of \(4.0\times10^6\,\text{m}\), while Planet B has a radius of \(6.0\times10^6\,\text{m}\).

Determine:

a) The ratio of their masses, \(M_B/M_A\).

b) The ratio of their surface escape speeds, \(v_B/v_A\).

Explain why the larger planet has the greater escape speed, even though increasing radius alone would reduce escape speed if mass remained constant.

[IMAGE: Two spherical planets of different sizes but identical average density, shown side by side. Label their radii R_A and R_B and indicate that their densities are equal. Make Planet B visibly larger. Do not include calculated mass or escape-speed ratios.]

### Solution 4

Planet B has 3.375 times the mass of Planet A and 1.50 times its escape speed.

The important detail is that the planets have equal average density. Increasing the radius therefore also increases the planet's mass.

**Step 1**: Use the equation for the volume of a sphere.

\[
V=\frac43\pi R^3
\]

Since average density is:

\[
\rho=\frac{M}{V}
\]

we can write:

\[
M=\rho\frac43\pi R^3
\]

Because the densities are equal:

\[
\frac{M_B}{M_A}
=
\left(\frac{R_B}{R_A}\right)^3
\]

**Step 2**: Substitute the radius values.

\[
\frac{M_B}{M_A}
=
\left(\frac{6.0}{4.0}\right)^3
\]

\[
=(1.5)^3
\]

\[
\boxed{\frac{M_B}{M_A}=3.375}
\]

Planet B has 3.375 times the mass of Planet A.

**Step 3**: Calculate the escape-speed ratio.

Use:

\[
\frac{v_B}{v_A}
=
\sqrt{
\frac{M_B}{M_A}
\frac{R_A}{R_B}
}
\]

Substitute:

\[
\frac{v_B}{v_A}
=
\sqrt{
3.375\times\frac{4.0}{6.0}
}
\]

\[
=\sqrt{2.25}
\]

\[
\boxed{\frac{v_B}{v_A}=1.50}
\]

Planet B has 50% greater escape speed.

The tempting prediction is that a larger radius must reduce escape speed. That would be correct if the planet's mass stayed constant.

However, equal density means that mass increases with the cube of the radius. This increase outweighs the radius effect, producing a higher escape speed.

### Question 5

A probe is launched vertically outwards from the surface of a spherical planet at \(0.80v_{\text{esc}}\), where \(v_{\text{esc}}\) is the planet's surface escape speed.

The probe travels outwards under gravity alone until it reaches a distance of \(2R\) from the planet's centre, where \(R\) is the planet's radius.

At this point, an instantaneous outward impulse changes the probe's speed to \(0.75v_{\text{esc}}\).

Assume the impulse acts in the direction of the probe's existing motion.

Determine:

a) The probe's speed immediately before the impulse, expressed as a fraction of the original surface escape speed.

b) The minimum increase in speed required at \(2R\) to make the probe escape.

c) Whether the probe escapes after receiving the stated impulse.

d) Its final speed at infinity.

[IMAGE: A spherical planet with a probe travelling vertically outwards from its surface towards a point twice the planet's radius from its centre. Label the surface radius R, the probe's starting speed as 0.80 v_esc, and the position of the instantaneous impulse at r = 2R. Show the supplied post-impulse speed of 0.75 v_esc. Keep the diagram to the given arrangement and do not show the unknown speed before the impulse or the local escape-speed threshold.]

### Solution 5

The probe does escape after the impulse. Its speed immediately before the impulse is approximately \(0.374v_{\text{esc}}\), and the minimum speed increase needed at \(2R\) is approximately \(0.333v_{\text{esc}}\). After the stated impulse, its speed at infinity is \(0.250v_{\text{esc}}\).

This question tests an important idea: escape speed depends on the distance from the planet's centre. The escape speed at \(2R\) is not the same as the escape speed at the surface.

**Step 1**: Calculate the probe's speed at \(2R\) before the impulse.

Use conservation of mechanical energy.

Initially:

\[
E_i=\frac12 m(0.80v_{\text{esc}})^2-\frac{GMm}{R}
\]

At \(2R\):

\[
E_f=\frac12 mv_2^2-\frac{GMm}{2R}
\]

Since mechanical energy is conserved:

\[
\frac12 m(0.80v_{\text{esc}})^2-\frac{GMm}{R}
=
\frac12 mv_2^2-\frac{GMm}{2R}
\]

Cancel \(m\) and rearrange:

\[
v_2^2=(0.80v_{\text{esc}})^2-\frac{GM}{R}
\]

We know:

\[
v_{\text{esc}}^2=\frac{2GM}{R}
\]

Therefore:

\[
\frac{GM}{R}=\frac12v_{\text{esc}}^2
\]

Substitute:

\[
v_2^2=0.64v_{\text{esc}}^2-0.50v_{\text{esc}}^2
\]

\[
v_2^2=0.14v_{\text{esc}}^2
\]

Taking the square root:

\[
\boxed{v_2=0.374v_{\text{esc}}}
\]

The probe is still moving outwards, but gravity has reduced its speed.

**Step 2**: Calculate the escape speed at \(2R\).

The escape-speed equation is:

\[
v_{\text{esc,local}}=\sqrt{\frac{2GM}{r}}
\]

At \(r=2R\):

\[
v_{\text{esc,local}}
=
\sqrt{\frac{2GM}{2R}}
\]

Simplifying:

\[
v_{\text{esc,local}}=\sqrt{\frac{GM}{R}}
\]

Using the original surface escape-speed equation:

\[
\boxed{v_{\text{esc,local}}=\frac{v_{\text{esc}}}{\sqrt2}}
\]

Therefore:

\[
v_{\text{esc,local}}=0.707v_{\text{esc}}
\]

**Step 3**: Find the minimum speed increase required.

The probe currently moves at \(0.374v_{\text{esc}}\).

To just escape from \(2R\), its speed must reach \(0.707v_{\text{esc}}\).

Therefore:

\[
\Delta v_{\min}=0.707v_{\text{esc}}-0.374v_{\text{esc}}
\]

\[
\boxed{\Delta v_{\min}=0.333v_{\text{esc}}}
\]

**Step 4**: Determine whether the stated impulse is sufficient.

The impulse increases the probe's speed to:

\[
v_{\text{after}}=0.75v_{\text{esc}}
\]

Compare this with the local escape speed:

\[
0.75v_{\text{esc}}>0.707v_{\text{esc}}
\]

Therefore, the probe escapes.

We can also confirm this using total mechanical energy per unit mass immediately after the impulse:

\[
\frac{E}{m}
=
\frac12(0.75v_{\text{esc}})^2-\frac{GM}{2R}
\]

Using \(GM/R=0.5v_{\text{esc}}^2\):

\[
\frac{E}{m}
=
0.28125v_{\text{esc}}^2
-
0.25v_{\text{esc}}^2
\]

\[
\frac{E}{m}=0.03125v_{\text{esc}}^2
\]

This is positive, confirming that the probe is unbound.

**Step 5**: Calculate the speed at infinity.

At infinity, gravitational potential energy approaches zero.

Conservation of energy gives:

\[
\frac12 mv_{\infty}^2=E
\]

Therefore:

\[
v_{\infty}^2=2\frac{E}{m}
\]

Substitute:

\[
v_{\infty}^2=0.0625v_{\text{esc}}^2
\]

Taking the square root:

\[
\boxed{v_{\infty}=0.250v_{\text{esc}}}
\]

The probe escapes with a non-zero speed because the impulse gave it more energy than the minimum required.

The tempting mistake is to compare its final speed of \(0.75v_{\text{esc}}\) with the original surface escape speed and conclude that it cannot escape.

But the probe is now at \(2R\), where the local escape speed is only \(0.707v_{\text{esc}}\).

This is the key distinction: always compare an object's current speed with the escape speed at its current distance from the planet's centre.

## Where to go next

Once you can calculate escape speed from planetary mass and radius, the next useful step is understanding how it compares with circular orbital speed.

Both quantities come from the same gravitational model, but they describe different types of motion.

Work through [Escape Velocity for HSC Physics: Derivation and Calculations](https://tutorgum.com/learn/resources/hsc-physics-escape-velocity/) for the orbital-speed connection, then use [How to Read Gravitational Energy Graphs for Escape Speed](https://tutorgum.com/learn/resources/how-to-read-gravitational-energy-graphs-for-escape-from-a-planet/) to practise interpreting bound and unbound motion.