Escape Velocity for HSC Physics: Derivation and Calculations

Throw a ball upwards and gravity brings it back. Throw it faster and it goes higher, but it still comes back. So here is the useful question: is there a speed high enough that the object never returns, even though gravity never actually switches off?

Before reading on, make a prediction. Does an escaping spacecraft need to reach some distance where gravity becomes zero?

It doesn’t. Gravity gets weaker with distance, but it remains non-zero at every finite distance. Escape velocity comes from a different idea: give the object enough kinetic energy at the start that it can climb out of the gravitational field without running out of energy.

01What does it mean to escape?

Picture gravity as an extremely long hill. Near a planet, the hill is steep. Far away, it becomes flatter and flatter, but there isn’t a sudden edge where the hill ends.

An object launched too slowly climbs partway up, loses speed, and may return. The minimum escape speed is the special case where it can keep moving farther and farther away, with its speed approaching zero only as its distance approaches infinity.

That analogy is useful, but it has a limit. A spacecraft isn’t literally rolling up a surface. The “height” of the gravitational hill is represented by gravitational potential energy.

For a mass \(m\) at distance \(r\) from the centre of a spherical body of mass \(M\),

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

where:

• \(U\) is gravitational potential energy in joules, J,
• \(G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}\) is the universal gravitational constant,
• \(M\) is the mass of the planet or other central body in kilograms,
• \(m\) is the mass of the spacecraft or projectile in kilograms, and
• \(r\) is the distance from the centre of the central body in metres.

The negative sign matters. We define gravitational potential energy to be zero infinitely far away, so an object near a planet has negative gravitational potential energy.

If that feels unfamiliar, review gravitational potential energy in a radial field before going further.

02Deriving the escape velocity equation

Suppose an object starts at distance \(r\) from a planet’s centre with speed \(v\).

Its initial mechanical energy is

\[
E_i=\frac{1}{2}mv^2-\frac{GMm}{r}
\]

Now consider the exact minimum speed needed to escape.

At an enormous distance from the planet:

• \(r\rightarrow\infty\), so gravitational potential energy approaches zero, and
• the object has only just escaped, so its speed also approaches zero.

Therefore the final mechanical energy for the minimum-speed case is

\[
E_f=0
\]

Conservation of mechanical energy gives

\[
\frac{1}{2}mv^2-\frac{GMm}{r}=0
\]

Rearrange:

\[
\frac{1}{2}mv^2=\frac{GMm}{r}
\]

The mass \(m\) of the escaping object cancels:

\[
\frac{1}{2}v^2=\frac{GM}{r}
\]

so

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

This is the escape velocity equation.

It is often better to call \(v_{\text{esc}}\) an escape speed, because the equation gives a magnitude rather than a direction.

03What the equation is really saying

The equation

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

contains three important ideas.

First, a more massive central body has a larger escape speed. A larger \(M\) means a deeper gravitational potential well.

Second, escape speed decreases as starting distance \(r\) increases. If you’re already far from the planet, less energy is needed to get away.

Third, the mass of the escaping object doesn’t appear at all.

A 1 kg probe and a 10 000 kg spacecraft need the same escape speed if they start from the same position, provided we ignore effects such as air resistance and propulsion efficiency. The heavier spacecraft needs more kinetic energy, but its gravitational potential energy also changes by a proportionally larger amount. The two mass factors cancel.

04Escape speed compared with orbital speed

There is a useful connection between escape velocity and circular orbital velocity.

For a circular orbit at radius \(r\),

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

while

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

Therefore,

\[
v_{\text{esc}}=\sqrt{2}\,v_{\text{orb}}
\]

at the same radius.

So a spacecraft already in circular orbit is not starting from zero. It already has substantial kinetic energy. To escape, it only needs its speed increased from \(v_{\text{orb}}\) to \(\sqrt{2}v_{\text{orb}}\).

This connection follows directly from the physics in orbital velocity.

05Escape velocity is not a speed you must keep

A common prediction is that an escaping spacecraft must remain above \(v_{\text{esc}}\) throughout its journey.

That isn’t how the equation works.

Escape velocity is the minimum speed required at a particular radius if the object then moves only under gravity. Once launched, the spacecraft slows as kinetic energy is converted into gravitational potential energy.

At the exact escape speed,

\[
v\rightarrow0
\]

as

\[
r\rightarrow\infty
\]

The spacecraft can therefore be moving below its original escape speed later and still be escaping.

There is another subtle point. The escape speed itself also becomes smaller as \(r\) increases:

\[
v_{\text{esc}}\propto\frac{1}{\sqrt r}
\]

So you should always compare an object’s current speed with the escape speed at its current radius, not with the escape speed at its starting position.

06What if the object is launched faster than escape velocity?

At the exact escape velocity, the object’s total mechanical energy is zero.

If

\[
v>v_{\text{esc}}
\]

then

\[
E>0
\]

The object escapes and still has kinetic energy remaining when it is extremely far away.

If its speed infinitely far away is \(v_\infty\), conservation of energy gives

\[
\frac{1}{2}mv^2-\frac{GMm}{r}
=
\frac{1}{2}mv_\infty^2
\]

Using \(v_{\text{esc}}^2=2GM/r\),

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

This relationship is useful in reverse as well. An asteroid arriving from far away can be accelerated by a planet’s gravity before reaching the surface.

Notice what you cannot do:

\[
v=v_{\text{esc}}+v_\infty
\]

Speeds do not simply add here, because the governing relationship comes from energy, which depends on \(v^2\).

07Does direction matter?

Suppose two probes start from the same radius with exactly the same speed, equal to the local escape speed. One is launched directly outwards. The other is launched sideways.

Ignoring collisions, atmosphere, other celestial bodies, and propulsion after launch, both have the same total mechanical energy:

\[
E=0
\]

Both are therefore at the boundary between bound and unbound motion.

Their paths are different because their directions, and therefore their angular momenta, are different. But escape is determined by the total energy.

This gives a useful decision rule:

“Bound” does not necessarily mean “falls straight back down”. A satellite in a stable elliptical or circular orbit also has negative total mechanical energy.

08Assumptions behind the escape velocity formula

The equation

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

is powerful because the model is simple. That also means you need to know its assumptions.

We assume:

• the central body can be treated as spherically symmetric,
• its mass \(M\) does not change,
• the escaping object’s mass is much smaller than \(M\),
• gravity from other bodies is ignored,
• there is no air resistance,
• no further propulsion acts after the initial speed is given, and
• mechanical energy is conserved.

For an HSC calculation, these assumptions are often built into the wording rather than stated explicitly.

09Questions and solutions

Question 1

The Moon has mass \(7.35\times10^{22}\ \text{kg}\) and radius \(1.74\times10^6\ \text{m}\).

Calculate the escape speed from the Moon’s surface.

Solution 1

The escape speed is approximately \(2.38\ \text{km s}^{-1}\).

Use

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

Substitute:

\[
v_{\text{esc}}
=
\sqrt{
\frac{
2(6.67\times10^{-11})(7.35\times10^{22})
}{
1.74\times10^6
}
}
\]

\[
v_{\text{esc}}
\approx2.38\times10^3\ \text{m s}^{-1}
\]

Therefore,

\[
v_{\text{esc}}\approx2.38\ \text{km s}^{-1}
\]

The Moon’s escape speed is much lower than Earth’s because the Moon has far less mass.

A tempting mistake is to use altitude above the Moon’s surface as \(r\). In this equation, \(r\) must be measured from the centre of the Moon.

Question 2

A satellite travels in a circular orbit of radius \(2R\) around a planet of mass \(M\).

Derive an expression for:

a. its circular orbital speed,

b. the escape speed at that radius, and

c. the minimum instantaneous increase in speed required to escape if the boost is applied in the direction of motion.

Solution 2

The required speed increase is

\[
\Delta v=(\sqrt2-1)\sqrt{\frac{GM}{2R}}
\]

For part a, circular orbital speed is

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

At \(r=2R\),

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

For part b,

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

so at \(r=2R\),

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

Alternatively,

\[
v_{\text{esc}}=\sqrt2\,v_{\text{orb}}
\]

For part c,

\[
\Delta v=v_{\text{esc}}-v_{\text{orb}}
\]

so

\[
\Delta v
=
\sqrt2\sqrt{\frac{GM}{2R}}
–
\sqrt{\frac{GM}{2R}}
\]

Therefore,

\[
\Delta v
=
(\sqrt2-1)\sqrt{\frac{GM}{2R}}
\]

The tempting route is to say the spacecraft needs the entire escape speed as an additional boost. It does not. Its existing orbital kinetic energy counts towards the energy required for escape.

Question 3

An asteroid approaches Earth from a very large distance with speed \(4.00\ \text{km s}^{-1}\).

Ignoring Earth’s atmosphere and the gravitational effects of other bodies, calculate the asteroid’s speed when it reaches Earth’s surface. Take Earth’s surface escape speed as \(11.2\ \text{km s}^{-1}\).

Explain why simply adding \(4.00\ \text{km s}^{-1}\) and \(11.2\ \text{km s}^{-1}\) is incorrect.

Solution 3

The asteroid reaches Earth’s surface at approximately \(11.9\ \text{km s}^{-1}\).

At a very large distance, Earth’s gravitational potential energy is approximately zero. The asteroid’s specific mechanical energy, meaning energy per kilogram, is therefore

\[
\frac{1}{2}v_\infty^2
\]

At Earth’s surface,

\[
\frac{1}{2}v^2-\frac{GM}{R}
=
\frac{1}{2}v_\infty^2
\]

Since

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

we obtain

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

Substitute the speeds in \(\text{km s}^{-1}\):

\[
v
=
\sqrt{(4.00)^2+(11.2)^2}
\]

\[
v
=
\sqrt{16.0+125.44}
=
\sqrt{141.44}
\]

\[
v\approx11.9\ \text{km s}^{-1}
\]

The speeds do not add directly because Earth’s gravity changes the asteroid’s kinetic energy. Kinetic energy depends on \(v^2\), so conservation of energy produces a sum of squared speeds.

The tempting answer,

\[
4.00+11.2=15.2\ \text{km s}^{-1},
\]

would give the asteroid much more energy than gravity has actually supplied.

Question 4

Planets A and B have the same surface gravitational field strength \(g\). Planet B has four times the radius of planet A.

a. Determine the ratio \(M_B/M_A\).

b. Determine the ratio of their surface escape speeds, \(v_{\text{esc},B}/v_{\text{esc},A}\).

c. A projectile is launched vertically from the surface of planet B at exactly the escape speed from the surface of planet A. Ignore atmospheres and rotation. Find the maximum distance of the projectile from the centre of planet B as a multiple of \(R_B\).

Solution 4

Planet B has 16 times the mass of planet A, twice the escape speed, and the projectile reaches a maximum distance of

\[
\frac{4}{3}R_B
\]

from B’s centre.

For part a, surface gravitational field strength is

\[
g=\frac{GM}{R^2}
\]

The planets have equal \(g\), so

\[
\frac{GM_A}{R_A^2}
=
\frac{GM_B}{R_B^2}
\]

Given

\[
R_B=4R_A,
\]

we have

\[
\frac{M_B}{M_A}
=
\frac{R_B^2}{R_A^2}
=
16
\]

For part b, escape speed is

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

Using \(GM=gR^2\),

\[
v_{\text{esc}}=\sqrt{2gR}
\]

Since both planets have the same \(g\),

\[
\frac{v_{\text{esc},B}}{v_{\text{esc},A}}
=
\sqrt{\frac{R_B}{R_A}}
=
\sqrt4
=
2
\]

So B requires twice the surface escape speed even though both planets have the same surface gravitational field strength.

That is the first trap: equal surface \(g\) does not mean equal escape speed. Escape depends on the entire gravitational potential well, not just the force per kilogram at one location.

For part c, let the launch speed from B be \(v_A\), where

\[
v_A^2=2gR_A
\]

The projectile’s initial kinetic energy per kilogram is

\[
\frac{1}{2}v_A^2=gR_A
\]

Planet B has

\[
R_B=4R_A
\]

and

\[
GM_B=gR_B^2
\]

The projectile’s initial specific mechanical energy at B’s surface is

\[
\varepsilon
=
gR_A-\frac{GM_B}{R_B}
\]

Since

\[
\frac{GM_B}{R_B}=gR_B=4gR_A,
\]

then

\[
\varepsilon=-3gR_A
\]

At maximum distance \(r_{\max}\), the projectile momentarily has zero speed, so

\[
\varepsilon=-\frac{GM_B}{r_{\max}}
\]

Thus

\[
-3gR_A
=
-\frac{gR_B^2}{r_{\max}}
\]

Using \(R_B=4R_A\),

\[
3gR_A
=
\frac{g(4R_A)^2}{r_{\max}}
\]

\[
3R_A=\frac{16R_A^2}{r_{\max}}
\]

Therefore,

\[
r_{\max}=\frac{16}{3}R_A
\]

Since \(R_B=4R_A\),

\[
r_{\max}
=
\frac{4}{3}R_B
\]

The projectile therefore rises only

\[
\frac{1}{3}R_B
\]

above B’s surface.

The deeper constraint is that surface gravitational field strength describes the local acceleration only. Planet B has the same surface \(g\), but that gravitational field extends across a much larger length scale.

Question 5

A spacecraft is launched radially outwards from the surface of a planet. The planet’s surface escape speed is \(10.0\ \text{km s}^{-1}\), and the spacecraft is initially launched at \(9.00\ \text{km s}^{-1}\).

Its engine can later provide one instantaneous increase in speed of exactly \(1.50\ \text{km s}^{-1}\), directed along its motion.

A student argues that the spacecraft should wait until it is far from the planet before firing the engine because the local escape speed will then be lower.

a. Determine whether firing the engine immediately at the surface allows the spacecraft to escape.

b. Without using the engine, determine the maximum radius the spacecraft would reach, expressed as a multiple of the planet’s radius \(R\).

c. Determine the greatest radius at which the \(1.50\ \text{km s}^{-1}\) engine boost can still be applied and produce escape.

d. Assess the student’s argument.

Solution 5

The spacecraft escapes if the boost is used early enough, but waiting too long makes the same \(1.50\ \text{km s}^{-1}\) boost insufficient. The latest usable point is approximately

\[
1.99R
\]

from the planet’s centre.

For part a, the spacecraft’s speed immediately after an engine burn at the surface would be

\[
v=9.00+1.50=10.50\ \text{km s}^{-1}
\]

The local escape speed is \(10.0\ \text{km s}^{-1}\), so

\[
10.50>10.0
\]

and the spacecraft escapes.

For part b, let the surface escape speed be \(V=10.0\ \text{km s}^{-1}\).

Since

\[
V^2=\frac{2GM}{R},
\]

we have

\[
\frac{GM}{R}=\frac{V^2}{2}
=
50.0\ \text{km}^2\text{s}^{-2}
\]

The spacecraft’s initial specific mechanical energy is

\[
\varepsilon
=
\frac{1}{2}v^2-\frac{GM}{R}
\]

\[
\varepsilon
=
\frac{1}{2}(9.00)^2-50.0
\]

\[
\varepsilon
=
40.5-50.0
=
-9.50\ \text{km}^2\text{s}^{-2}
\]

At the maximum radius, the speed is zero, so

\[
\varepsilon=-\frac{GM}{r_{\max}}
\]

Therefore,

\[
-9.50
=
-\frac{50.0R}{r_{\max}}
\]

which gives

\[
r_{\max}
=
\frac{50.0}{9.50}R
\]

\[
r_{\max}\approx5.26R
\]

The spacecraft could therefore coast surprisingly far from the surface without escaping.

For part c, the spacecraft begins with a specific-energy deficit of

\[
9.50\ \text{km}^2\text{s}^{-2}
\]

relative to escape.

Suppose its speed immediately before the engine burn is \(v\). Increasing its speed by

\[
\Delta v=1.50\ \text{km s}^{-1}
\]

changes its kinetic energy per kilogram by

\[
\Delta\varepsilon
=
\frac{1}{2}(v+\Delta v)^2-\frac{1}{2}v^2
\]

Expand:

\[
\Delta\varepsilon
=
v\Delta v+\frac{1}{2}(\Delta v)^2
\]

For the spacecraft to just escape,

\[
v(1.50)+\frac{1}{2}(1.50)^2=9.50
\]

\[
1.50v+1.125=9.50
\]

\[
v=5.58\ \text{km s}^{-1}
\]

We now need to find where the coasting spacecraft has slowed to this speed.

At radius \(r\),

\[
\frac{1}{2}v^2-\frac{GM}{r}=-9.50
\]

Using \(GM=50.0R\) in \(\text{km}^2\text{s}^{-2}\) units,

\[
\frac{1}{2}(5.58)^2-\frac{50.0R}{r}=-9.50
\]

\[
15.6-\frac{50.0R}{r}=-9.50
\]

\[
\frac{50.0R}{r}=25.1
\]

Therefore,

\[
r\approx1.99R
\]

So the engine must fire at approximately \(1.99R\) or closer to the planet’s centre.

For part d, the student’s argument misses an important interaction.

It is true that local escape speed decreases with increasing radius. That part of the argument is correct.

But the spacecraft is also slowing down as it climbs. More importantly, a fixed increase in speed \(\Delta v\) gives a kinetic-energy increase of

\[
\Delta K
=
m\left(v\Delta v+\frac{1}{2}\Delta v^2\right)
\]

The same \(1.50\ \text{km s}^{-1}\) boost therefore adds more kinetic energy when the spacecraft is already moving quickly.

At the surface, the spacecraft is moving at \(9.00\ \text{km s}^{-1}\), so the burn easily produces escape. If it waits until its maximum radius of \(5.26R\), its speed has fallen to zero. The same burn would then give it only \(1.50\ \text{km s}^{-1}\), far below the local escape speed there.

The tempting idea was to compare only the changing escape speed. The missing constraint was the spacecraft’s own changing speed, and therefore the amount of energy supplied by a fixed \(\Delta v\).

10Where escape velocity leads next

Escape velocity is really an energy result disguised as a speed equation. Once you can move comfortably between

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

\[
v_{\text{orb}}=\sqrt{\frac{GM}{r}},
\]

and

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

orbital questions stop looking like unrelated formulas.

They become different energy states in the same gravitational field: negative total energy for bound motion, zero energy at the escape boundary, and positive energy for unbound motion. That is the useful connection to carry into circular orbits, elliptical trajectories, and spacecraft energy changes.