How to Calculate Speed Remaining After Escaping a Planet

A probe is launched from Earth’s surface at 15.0 km/s. Earth’s escape speed at the surface is approximately 11.2 km/s.

The probe escapes Earth, but does it still travel at 15.0 km/s once it is infinitely far away? No. Earth’s gravity slows the probe as it moves away, converting some of its kinetic energy into gravitational potential energy.

However, the probe started faster than escape speed, so it does not lose all its kinetic energy. It retains a final speed of approximately 9.99 km/s at infinity.

This remaining speed is called the speed at infinity. It is calculated using:

\[ \boxed{v_\infty=\sqrt{v_0^2-\frac{2GM}{r_0}}} \]

Here, \(v_0\) is the initial speed, \(M\) is the planet’s mass, \(G\) is the universal gravitational constant, and \(r_0\) is the initial distance from the planet’s centre.

The important condition is that the initial speed must be greater than the escape speed at that starting position.

01Why you cannot simply subtract the escape speed

Suppose a spacecraft launches at 15.0 km/s and Earth’s escape speed is 11.2 km/s.

You might predict that its remaining speed is:

\[ 15.0-11.2=3.8\ \mathrm{km/s} \]

This seems reasonable, but it is incorrect.

Escape speed is not a quantity of speed that gets used up. It is the minimum initial speed required to reach infinity with zero final speed.

The difference comes from kinetic energy:

\[ E_k=\frac12mv^2 \]

Kinetic energy depends on the square of velocity, not velocity itself.

When a spacecraft escapes a planet, its initial kinetic energy is divided between:

• The gravitational potential energy it gains while moving away.
• The kinetic energy it retains at infinity.

Therefore, we must subtract the squared speeds before taking the square root.

02The physical idea: Newton’s cannonball

There is a useful historical thought experiment behind this problem.

Imagine Newton’s cannonball: a cannon positioned on top of a very high mountain, pointing horizontally away from the mountain.

What happens if you fire a cannonball?

At a low launch speed, it travels a short distance before gravity pulls it into the ground. Increase the launch speed, and it travels further before reaching Earth’s surface.

But what happens if the launch speed becomes sufficiently large?

The cannonball can keep falling towards Earth without ever reaching the ground. It follows a curved path around the planet.

This thought experiment helped establish an important idea: orbital motion and falling motion are not separate processes. Both are consequences of gravity continuously changing an object’s velocity.

It also raises the question we need to answer: when does a projectile stop being gravitationally bound to Earth?

Newton’s imagined cannon is a theoretical model, not an experiment that measured escape speed. It assumes no atmospheric resistance and an idealised spherical Earth.

A real launch would need to account for atmospheric drag, Earth’s rotation, and other gravitational influences. For the HSC calculation, we isolate gravity and use conservation of mechanical energy.

03The energy calculation

To calculate the remaining speed, we need to understand what happens to the object’s energy.

We choose gravitational potential energy to be zero at infinity:

\[ U(\infty)=0 \]

At a finite distance \(r\) from the centre of a spherical planet, gravitational potential energy is:

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

This negative value means that the object is gravitationally bound to the planet.

If you need a more detailed explanation of this sign convention, see Why Gravitational Potential Energy Is Negative Around a Planet.

What happens to the energy as the object escapes?

Consider a spacecraft launched outwards.

Initially, it has:

• Positive kinetic energy because it is moving.
• Negative gravitational potential energy because it is gravitationally bound to the planet.

As it moves further away, gravitational potential energy increases towards zero.

At the same time, kinetic energy decreases because gravity acts against its outward motion.

The total mechanical energy remains constant, provided no external forces do work on the spacecraft.

The conservation of mechanical energy equation is:

\[ E_i=E_f \]

At the initial position:

\[ E_i=E_{k,i}+U_i \]

At infinity:

\[ E_f=E_{k,f}+U_f \]

Substituting the gravitational potential energy equation:

\[ \frac12mv_0^2-\frac{GMm}{r_0} = \frac12mv_\infty^2+0 \]

Notice that the final gravitational potential energy is zero because we have chosen infinity as our reference point.

Rearranging:

\[ \frac12mv_0^2-\frac{GMm}{r_0} = \frac12mv_\infty^2 \]

The object’s mass appears on both sides, so it cancels.

Multiplying by 2 and rearranging gives:

\[ v_0^2-\frac{2GM}{r_0}=v_\infty^2 \]

Therefore:

\[ \boxed{v_\infty=\sqrt{v_0^2-\frac{2GM}{r_0}}} \]

This is the equation you need.

Notice that the escaping object’s mass does not appear in the final expression. If you want to understand that result separately, read Does Escape Speed Depend on Mass? HSC Physics Explained.

What each variable represents

Use \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\) unless the question supplies a different value.

One particularly important detail is \(r_0\).

It is measured from the planet’s centre, not its surface.

If the spacecraft launches from an altitude \(h\) above the surface, then:

\[ r_0=R+h \]

where \(R\) is the planet’s radius.

04The three possible outcomes

Before using the equation, check whether the object actually has enough energy to escape.

The total mechanical energy is:

\[ E=E_k+U \]

Its sign determines the outcome.

This distinction is important.

A negative total mechanical energy does not mean negative kinetic energy. Kinetic energy cannot be negative.

It means the object is gravitationally bound. The magnitude of its negative total energy represents the additional energy required to bring it to the marginal escape condition.

For an object that starts with positive total mechanical energy, its remaining kinetic energy at infinity is exactly equal to its initial total mechanical energy.

05A reliable method for HSC calculations

When a question asks you to calculate the speed remaining after escaping a planet, use this sequence.

1. Identify the initial speed, \(v_0\).
2. Find the initial distance from the planet’s centre, \(r_0\). Add the planet’s radius to the altitude if necessary.
3. Identify the planet’s mass, \(M\), and gravitational constant, \(G\).
4. Check that the initial speed exceeds the local escape speed.
5. Substitute into the speed-at-infinity equation.
6. Calculate the final speed and express it in the correct units.

You can also calculate the escape speed at the starting position:

\[ v_{\mathrm{esc}}=\sqrt{\frac{2GM}{r_0}} \]

This gives a convenient alternative form:

\[ \boxed{v_\infty=\sqrt{v_0^2-v_{\mathrm{esc}}^2}} \]

Use this form when the question gives you the escape speed directly.

06What if the initial speed is below escape speed?

Now consider a spacecraft launched from Earth’s surface at only \(10.0\ \mathrm{km/s}\).

Earth’s escape speed at the surface is approximately \(11.2\ \mathrm{km/s}\).

The spacecraft does not have enough energy to escape.

If you substitute these values into the speed-at-infinity equation, you get:

\[ v_\infty=\sqrt{10.0^2-11.2^2} \]

\[ v_\infty=\sqrt{-25.44} \]

A negative number appears inside the square root.

This does not mean the spacecraft has an imaginary physical speed. It means the spacecraft cannot reach infinity under the stated assumptions.

We can confirm this by calculating its initial total mechanical energy per unit mass:

\[ \frac{E}{m}= \frac12v_0^2-\frac{GM}{R} \]

Using SI units:

\[ \frac{E}{m}= \frac12(1.00\times10^4)^2 -\frac{3.99\times10^{14}}{6.37\times10^6} \]

\[ \frac{E}{m} =5.00\times10^7-6.26\times10^7 \]

\[ \frac{E}{m}=-1.26\times10^7\ \mathrm{J/kg} \]

The total mechanical energy is negative, confirming that the spacecraft remains gravitationally bound.

If it launches radially outwards, it eventually reaches a maximum distance from Earth, momentarily stops, and falls back.

At this maximum distance, its kinetic energy is zero.

This is why you should always check the sign of the total mechanical energy before interpreting a calculated speed.

A negative total energy means the object is bound. A positive total energy means it can escape with kinetic energy remaining.

07Questions and solutions

Try these questions without looking at the solutions first. They progress from standard escape-speed calculations to more involved energy reasoning.

Question 1

A small object is launched radially outwards from the surface of a spherical planet.

Its initial speed is \(13.0\ \mathrm{km/s}\), and the escape speed at the surface is \(10.0\ \mathrm{km/s}\).

Calculate the speed of the object at infinity.

Solution 1

The speed at infinity is \(8.31\ \mathrm{km/s}\).

We use the difference between the squared initial speed and squared escape speed:

\[ v_\infty=\sqrt{v_0^2-v_{\mathrm{esc}}^2} \]

Substitute:

\[ v_\infty=\sqrt{13.0^2-10.0^2} \]

\[ v_\infty=\sqrt{169-100} \]

\[ v_\infty=\sqrt{69} \]

Therefore:

\[ \boxed{v_\infty=8.31\ \mathrm{km/s}} \]

The object has enough initial kinetic energy to escape, but gravity converts some of that energy into gravitational potential energy during its outward journey.

Question 2

A \(250\ \mathrm{kg}\) spacecraft is launched from a position where the local escape speed is \(10.0\ \mathrm{km/s}\).

Its initial speed is \(12.5\ \mathrm{km/s}\).

Calculate:

a) Its speed at infinity.

b) Its kinetic energy at infinity.

c) The percentage of its initial kinetic energy that remains at infinity.

Explain what would happen to its speed at infinity if its mass were doubled, while all other conditions remained unchanged.

Solution 2

The spacecraft reaches infinity with a speed of \(7.50\ \mathrm{km/s}\), retaining 36.0% of its initial kinetic energy.

Part a: Calculate the speed at infinity.

Use:

\[ v_\infty=\sqrt{v_0^2-v_{\mathrm{esc}}^2} \]

Substitute:

\[ v_\infty=\sqrt{12.5^2-10.0^2} \]

\[ v_\infty=\sqrt{156.25-100} \]

\[ v_\infty=\sqrt{56.25} \]

Therefore:

\[ \boxed{v_\infty=7.50\ \mathrm{km/s}} \]

Part b: Calculate the final kinetic energy.

Convert the final speed into metres per second:

\[ v_\infty=7.50\times10^3\ \mathrm{m/s} \]

Use the kinetic energy equation:

\[ E_{k,\infty}=\frac12mv_\infty^2 \]

Substitute:

\[ E_{k,\infty} =\frac12(250)(7.50\times10^3)^2 \]

\[ \boxed{E_{k,\infty}=7.03\times10^9\ \mathrm{J}} \]

Part c: Calculate the percentage of initial kinetic energy retained.

First, calculate the initial kinetic energy:

\[ E_{k,i}=\frac12mv_0^2 \]

\[ E_{k,i}=\frac12(250)(12.5\times10^3)^2 \]

\[ E_{k,i}=1.95\times10^{10}\ \mathrm{J} \]

The percentage remaining is:

\[ \text{Percentage}= \frac{E_{k,\infty}}{E_{k,i}}\times100 \]

\[ =\frac{7.03\times10^9}{1.95\times10^{10}}\times100 \]

\[ \boxed{\text{Percentage}=36.0\%} \]

The remaining kinetic energy is 36.0% of the initial kinetic energy. The rest has been converted into gravitational potential energy.

What happens if the spacecraft’s mass doubles?

Its speed at infinity remains \(7.50\ \mathrm{km/s}\).

This is because the mass of the spacecraft cancels when we apply conservation of mechanical energy. However, its final kinetic energy doubles because kinetic energy is directly proportional to mass when speed is unchanged.

Question 3

A spacecraft launches from Earth’s surface at \(13.0\ \mathrm{km/s}\). Earth’s escape speed at the surface is \(11.2\ \mathrm{km/s}\).

As it moves outwards, its speed decreases. At a distance of \(2R\) from Earth’s centre, its speed is approximately \(10.3\ \mathrm{km/s}\).

A student argues:

“The spacecraft has slowed below Earth’s escape speed of 11.2 km/s, so it must now be gravitationally bound again.”

Explain why this conclusion is incorrect.

Calculate the local escape speed at \(2R\) and determine the spacecraft’s speed at infinity.

Solution 3

The spacecraft remains unbound and escapes Earth. The student’s mistake is comparing the spacecraft’s current speed with the escape speed at its original position.

Escape speed depends on distance from the planet’s centre.

At a greater distance, less additional energy is required to reach infinity, so the local escape speed is smaller.

Step 1: Calculate the local escape speed at \(2R\).

At the surface:

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

At twice the radius:

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

Therefore:

\[ v_{\mathrm{esc},2R} =\frac{v_{\mathrm{esc},R}}{\sqrt2} \]

Substitute:

\[ v_{\mathrm{esc},2R}= \frac{11.2}{\sqrt2} \]

\[ \boxed{v_{\mathrm{esc},2R}=7.92\ \mathrm{km/s}} \]

The spacecraft’s current speed is 10.3 km/s, which is greater than the local escape speed.

It is still unbound.

Step 2: Calculate the speed at infinity.

We can use the initial conditions at Earth’s surface:

\[ v_\infty=\sqrt{v_0^2-v_{\mathrm{esc},R}^2} \]

\[ v_\infty=\sqrt{13.0^2-11.2^2} \]

\[ v_\infty=\sqrt{169-125.44} \]

\[ v_\infty=\sqrt{43.56} \]

Therefore:

\[ \boxed{v_\infty=6.60\ \mathrm{km/s}} \]

The spacecraft continues to slow as it travels away, but its total mechanical energy remains positive.

The key lesson is that escape speed is not a fixed value for an entire journey. It changes with distance from the planet.

Question 4

A spacecraft travels radially outwards from Earth’s surface without using any engines after launch.

When it reaches a distance of \(3R\) from Earth’s centre, its speed is measured to be \(6.00\ \mathrm{km/s}\).

Determine:

a) Its original launch speed from Earth’s surface.

b) Whether the spacecraft will escape Earth.

Use:

• \(GM=3.99\times10^{14}\ \mathrm{m^3/s^2}\)
• \(R=6.37\times10^6\ \mathrm{m}\)

Assume no atmospheric resistance or other gravitational influences.

Solution 4

The original launch speed was approximately \(10.9\ \mathrm{km/s}\), so the spacecraft does not escape Earth.

We can work backwards using conservation of mechanical energy.

Part a: Determine the original launch speed.

Step 1: Calculate the total mechanical energy per unit mass at the measured position.

The spacecraft is at a distance of \(3R\), not an altitude of \(3R\).

Its total mechanical energy per unit mass is:

\[ \frac{E}{m}= \frac12v^2-\frac{GM}{r} \]

Substitute:

\[ \frac{E}{m} = \frac12(6.00\times10^3)^2 - \frac{3.99\times10^{14}} {3(6.37\times10^6)} \]

\[ \frac{E}{m} = 1.80\times10^7 - 2.09\times10^7 \]

\[ \frac{E}{m}=-2.88\times10^6\ \mathrm{J/kg} \]

Step 2: Apply conservation of mechanical energy at Earth’s surface.

At the surface:

\[ \frac{E}{m} = \frac12v_0^2-\frac{GM}{R} \]

Equate the initial and final expressions:

\[ \frac12v_0^2-\frac{GM}{R} =-2.88\times10^6 \]

Rearrange:

\[ \frac12v_0^2 = \frac{GM}{R}-2.88\times10^6 \]

Substitute:

\[ \frac12v_0^2 = 6.26\times10^7-2.88\times10^6 \]

\[ v_0^2=1.20\times10^8 \]

Taking the square root:

\[ \boxed{v_0=10.9\ \mathrm{km/s}} \]

Part b: Determine whether it escapes.

Earth’s escape speed at the surface is:

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

Substitute:

\[ v_{\mathrm{esc}} = \sqrt{ \frac{2(3.99\times10^{14})} {6.37\times10^6} } \]

\[ v_{\mathrm{esc}}=11.2\ \mathrm{km/s} \]

The original launch speed was only 10.9 km/s.

Therefore:

\[ 10.9<11.2 \]

The spacecraft does not escape.

Its negative total mechanical energy confirms that it remains gravitationally bound and will eventually stop moving outwards before falling back towards Earth.

This question demonstrates an important technique: you do not always need to know the initial speed. If the final position and speed are known, conservation of energy allows you to reconstruct the earlier state.

Question 5

A spacecraft is launched radially outwards from Earth’s surface at exactly the escape speed.

When it reaches a distance of \(2R\) from Earth’s centre, its speed is \(7.92\ \mathrm{km/s}\).

At this position, it receives an instantaneous thruster impulse that changes its velocity by \(3.00\ \mathrm{km/s}\).

Consider two possible situations:

• Case A: The change in velocity is directed parallel to the spacecraft’s existing outward velocity.
• Case B: The change in velocity is directed perpendicular to the spacecraft’s existing outward velocity.

After the impulse, the spacecraft moves under gravity alone.

For each case, calculate the spacecraft’s speed at infinity.

Explain why the same magnitude of velocity change produces different final speeds.

Solution 5

The spacecraft retains 7.52 km/s in Case A and 3.00 km/s in Case B at infinity.

Although both thruster impulses have the same magnitude, they produce different changes in kinetic energy because their directions are different.

This is the important part of the question.

Case A: The impulse is parallel to the existing velocity.

Step 1: Calculate the new speed immediately after the impulse.

Both velocities point outwards in the same direction, so their magnitudes add:

\[ v_{\mathrm{new}}=7.92+3.00 \]

\[ v_{\mathrm{new}}=10.92\ \mathrm{km/s} \]

Step 2: Calculate the speed at infinity.

The local escape speed at \(2R\) is \(7.92\ \mathrm{km/s}\).

Therefore:

\[ v_\infty= \sqrt{v_{\mathrm{new}}^2-v_{\mathrm{esc},2R}^2} \]

Substitute:

\[ v_\infty= \sqrt{10.92^2-7.92^2} \]

\[ v_\infty=\sqrt{119.2464-62.7264} \]

\[ v_\infty=\sqrt{56.52} \]

Therefore:

\[ \boxed{v_\infty=7.52\ \mathrm{km/s}} \]

Case B: The impulse is perpendicular to the existing velocity.

Step 1: Calculate the new speed immediately after the impulse.

The velocity change is perpendicular to the original velocity.

We therefore use Pythagoras’ theorem:

\[ v_{\mathrm{new}}^2=7.92^2+3.00^2 \]

\[ v_{\mathrm{new}}^2=62.7264+9.00 \]

\[ v_{\mathrm{new}}^2=71.7264 \]

Step 2: Calculate the speed at infinity.

Again, use the local escape speed at \(2R\):

\[ v_\infty^2= v_{\mathrm{new}}^2-v_{\mathrm{esc},2R}^2 \]

Substitute:

\[ v_\infty^2=71.7264-62.7264 \]

\[ v_\infty^2=9.00 \]

Therefore:

\[ \boxed{v_\infty=3.00\ \mathrm{km/s}} \]

Why are the answers different?

The tempting assumption is that the same change in velocity must produce the same increase in kinetic energy.

That is not true.

Kinetic energy depends on the magnitude of the final velocity, which depends on the angle between the original velocity and the velocity change.

The change in kinetic energy is:

\[ \Delta E_k= \frac12m \left( |\vec v+\Delta\vec v|^2-v^2 \right) \]

Expanding:

\[ \Delta E_k= m\vec v\cdot\Delta\vec v +\frac12m(\Delta v)^2 \]

In Case A, the velocity change is parallel to the original velocity. The dot product is positive, so the kinetic energy increases by a relatively large amount.

In Case B, the velocity change is perpendicular to the original velocity. The dot product is zero, so the kinetic energy increases by only \(\frac12m(\Delta v)^2\).

Both spacecraft escape because both impulses increase the initially zero total mechanical energy to a positive value.

However, Case A gives the spacecraft considerably more kinetic energy at infinity.

The important lesson is that an energy calculation depends on the actual change in kinetic energy, not simply the magnitude of a velocity change.

08Where to go next

Calculating the speed remaining at infinity is one application of conservation of mechanical energy in gravitational fields.

The next useful step is understanding how energy behaves when a spacecraft remains in orbit rather than escaping.

A circular satellite orbit has a different energy relationship because its kinetic energy is linked to its orbital radius through the centripetal force provided by gravity.

In particular, the total mechanical energy of a circular orbit is:

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

This explains why a satellite can move more slowly in a higher circular orbit while having greater total mechanical energy.

Keep the distinction clear:

• For an escaping object, compare its initial energy with the energy remaining at infinity.
• For a satellite changing circular orbits, compare the total mechanical energies of its initial and final orbits.

Both problems use conservation of mechanical energy, but they describe different types of motion.