Why Escape Does Not Mean Gravity Has Become Zero

A spacecraft can escape a planet while the planet’s gravity is still pulling on it. There is no distance at which gravity suddenly switches off and allows the spacecraft to leave. Escape means the spacecraft is no longer gravitationally bound, not that the gravitational field has become zero.

The distinction comes down to energy. We define gravitational potential energy to be zero at infinity. At the minimum escape energy, the spacecraft’s total mechanical energy is zero, even though its kinetic energy is positive and its gravitational potential energy is negative at every finite distance.

Before looking at the equations, consider this: a spacecraft escapes Earth and travels to a distance of four Earth radii from Earth’s centre. Has Earth’s gravitational field disappeared?

No. Its gravitational field strength is still one-sixteenth of its value at Earth’s surface. The spacecraft is escaping despite gravity remaining present.

01Newton’s cannon: escaping without switching gravity off

There is a useful historical thought experiment that helps explain this.

Isaac Newton imagined a cannon positioned on top of an extremely high mountain. The cannon fires a projectile horizontally, and we observe what happens as its launch speed increases.

Start with a relatively slow projectile. It travels forwards, but gravity pulls it downwards until it hits Earth’s surface.

Now increase its launch speed. It travels further before hitting the surface.

Increase the speed enough, and something interesting happens: the projectile continually falls towards Earth but keeps missing it. It enters a circular orbit.

Increase the speed beyond the escape threshold, and the projectile follows an open trajectory rather than a closed orbit. It does not return.

Notice what has changed.

The projectile’s initial speed determines its trajectory. Gravity continues acting throughout the journey.

Newton’s cannon is a thought experiment, not an actual cannon experiment. It deliberately ignores atmospheric resistance and assumes an idealised spherical Earth. Its value is that it separates two ideas that are easy to confuse:

• The trajectory describes how an object moves.
• The gravitational force determines how its motion changes.

An escaping object does not need gravity to disappear. It needs enough energy to avoid returning.

The energy approach gives us a much more precise way to understand this result.

02Why gravitational potential energy is negative

For a spherical planet, the gravitational potential energy of an object at distance \(r\) from the planet’s centre is

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

where:

• \(U\) is gravitational potential energy, measured 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, in kilograms.
• \(m\) is the mass of the object, in kilograms.
• \(r\) is the distance between their centres of mass, in metres.

The negative sign is important.

We choose gravitational potential energy to be zero at infinity:

\[
U(\infty)=0
\]

This is our reference point. It means that at any finite distance from the planet, gravitational potential energy is negative.

But why?

Imagine moving a spacecraft slowly away from a planet, starting at a distance \(r\), until it is infinitely far away.

You must supply energy to move it against gravity. Gravity is pulling inwards, while your applied force must do work outwards.

The change in gravitational potential energy is

\[
\begin{aligned}
\Delta U &= U_f-U_i\\
&=0-\left(-\frac{GMm}{r}\right)\\
&=\frac{GMm}{r}
\end{aligned}
\]

The spacecraft gains positive gravitational potential energy as it moves away.

Its initial potential energy was negative because it was below the chosen zero-energy reference.

Think of gravitational potential energy like height measured relative to sea level. A location at zero metres elevation is not necessarily flat. It simply sits at the chosen reference height.

Similarly, zero gravitational potential energy does not mean zero gravitational force.

The analogy has a limit: gravitational potential is an energy property of a three-dimensional field, not a literal physical hill.

03Gravitational potential energy and gravitational field strength are different

This is the distinction that answers the original question.

Gravitational potential energy describes the energy associated with an object’s position in a gravitational field.

Gravitational field strength describes the gravitational force per unit mass at a particular location.

For a spherical planet, the magnitude of gravitational field strength is

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

Its direction is always towards the planet’s centre.

Compare the two equations:

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

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

Notice that they depend on distance differently.

• Gravitational potential energy varies with \(1/r\).
• Gravitational field strength varies with \(1/r^2\).

As distance increases, both approach zero. However, neither is exactly zero at any finite distance.

For example, suppose an object moves from the planet’s surface to four times the planet’s radius from its centre.

Its gravitational field strength becomes

\[
g_{4R}=\frac{g_s}{4^2}
\]

\[
g_{4R}=\frac{g_s}{16}
\]

Here, \(g_s\) is the gravitational field strength at the surface, and \(R\) is the planet’s radius.

The gravitational field is now much weaker, but it is certainly not zero.

This is why you should never use gravitational potential energy as a direct measurement of gravitational field strength. They describe related aspects of gravity, but they are not interchangeable.

04What zero total mechanical energy actually means

Now consider a spacecraft travelling away from a planet.

Its total mechanical energy is the sum of its kinetic energy and gravitational potential energy:

\[
E=K+U
\]

Since kinetic energy is

\[
K=\frac12mv^2
\]

we can write

\[
E=\frac12mv^2-\frac{GMm}{r}
\]

For an isolated gravitational system, total mechanical energy is conserved.

This means the total remains constant as the spacecraft moves, even though kinetic energy and gravitational potential energy change.

For an object travelling at exactly the minimum escape speed, its total mechanical energy is zero:

\[
E=0
\]

Therefore,

\[
K+U=0
\]

Rearranging:

\[
K=-U
\]

Substituting the expressions for kinetic and gravitational potential energy gives

\[
\frac12mv^2=\frac{GMm}{r}
\]

The important result is that the kinetic energy is positive and the gravitational potential energy is negative. They cancel to give zero total mechanical energy.

Zero total energy does not mean zero kinetic energy.

At any finite distance, the spacecraft still has kinetic energy and is still experiencing gravitational attraction.

What happens as the spacecraft travels further?

For an object on the exact escape trajectory, its kinetic energy continually decreases as it moves away from the planet.

Its gravitational potential energy becomes less negative, approaching zero.

The two changes balance perfectly.

Consider the following illustrative values for an ideal spacecraft with zero total mechanical energy.

The energy columns are expressed as multiples of \(GMm/R\). The field-strength column is expressed as a multiple of the surface field strength.

Notice two things.

First, the positive and negative energy contributions always cancel.

Second, the gravitational field becomes weaker, but it remains non-zero at every listed distance.

Only in the limit as \(r\) approaches infinity do both gravitational potential energy and the spacecraft’s kinetic energy approach zero.

Infinity is a mathematical limit, not a finite boundary that the spacecraft crosses.

This is how an object can escape while gravity is still acting on it.

05Three possible energy states

The sign of total mechanical energy tells us whether an object is gravitationally bound, provided we use the convention \(U(\infty)=0\).

There is an important detail about negative energy.

A negative total mechanical energy does not necessarily mean an object will fall directly back to the planet.

A satellite in a circular or elliptical orbit also has negative total mechanical energy. It remains gravitationally bound, but its sideways motion allows it to keep orbiting.

The distinction is between bound and unbound motion, not simply moving towards or away from a planet.

If you need to revisit how the minimum escape speed is derived, see Escape Velocity for HSC Physics: Derivation and Calculations.

08A reliable method for HSC questions

When a question involves escape, gravitational potential energy, or the boundary between bound and unbound motion, use this sequence.

1. Identify the reference point. Unless stated otherwise, take gravitational potential energy to be zero at infinity.

2. Calculate the potential energy. Use \(U=-GMm/r\), measuring \(r\) from the planet’s centre.

3. Calculate total mechanical energy. Add kinetic and gravitational potential energy:

\[
E=\frac12mv^2-\frac{GMm}{r}
\]

1. Interpret the sign. Negative energy means bound, zero means minimum-energy escape, and positive energy means unbound motion.

2. Calculate gravitational field strength separately if required. Use \(g=GM/r^2\). Never assume it is zero just because total mechanical energy is zero.

The last step is particularly important. Total energy and gravitational field strength answer different physical questions.

09Questions and solutions

Question 1

A \(400\ \text{kg}\) probe is travelling away from a planet on an exact escape trajectory.

At a distance of \(2R\) from the planet’s centre, the value of \(GM/R\) is \(3.00\times10^7\ \text{J kg}^{-1}\).

Calculate:

a. The probe’s gravitational potential energy.

b. Its kinetic energy.

c. Its total mechanical energy.

d. The gravitational field strength as a fraction of the surface field strength.

Solution 1

The probe has gravitational potential energy of \(-6.00\times10^9\ \text{J}\), kinetic energy of \(+6.00\times10^9\ \text{J}\), and zero total mechanical energy. The gravitational field strength is one-quarter of its surface value.

Step 1: Calculate gravitational potential energy.

Use

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

Since \(r=2R\),

\[
U=-\frac{GMm}{2R}
\]

Substitute:

\[
U=-\frac{(3.00\times10^7)(400)}{2}
\]

\[
U=-6.00\times10^9\ \text{J}
\]

Step 2: Calculate kinetic energy.

The probe is on an exact escape trajectory, so

\[
E=0
\]

Using

\[
E=K+U
\]

we obtain

\[
K=-U
\]

Therefore,

\[
K=+6.00\times10^9\ \text{J}
\]

Step 3: State the total mechanical energy.

\[
E=K+U
\]

\[
E=6.00\times10^9-6.00\times10^9
\]

\[
E=0\ \text{J}
\]

Step 4: Calculate the gravitational field strength ratio.

Gravitational field strength varies inversely with the square of the distance:

\[
g\propto\frac{1}{r^2}
\]

Therefore,

\[
\frac{g_{2R}}{g_s}=\frac{R^2}{(2R)^2}
\]

\[
\frac{g_{2R}}{g_s}=\frac14
\]

The gravitational field strength is one-quarter of its surface value.

The important point is that the probe has zero total mechanical energy, but its kinetic energy and gravitational potential energy are both non-zero.

Question 2

A probe is launched vertically outwards from the surface of a planet.

It travels outwards until it momentarily stops at a distance of \(5R\) from the planet’s centre, where \(R\) is the planet’s radius.

Assume:

• The probe moves under gravity alone.
• Its initial and final motion is radial.
• The planet’s mass remains constant.

Determine:

a. The initial launch speed as a fraction of the surface escape speed.

b. The gravitational field strength at the maximum distance as a fraction of the surface field strength.

Explain why the probe does not escape, even though it travels five planetary radii from the centre.

Solution 2

The initial speed is approximately \(0.894\) times the surface escape speed, and the gravitational field strength at the turning point is \(1/25\) of its surface value. The probe does not escape because its total mechanical energy is negative.

Step 1: Consider the energy at the maximum distance.

At the turning point, the probe’s instantaneous speed is zero.

Therefore,

\[
E=-\frac{GMm}{5R}
\]

Step 2: Express the same energy at the surface.

Let the initial speed be \(v_0\).

At the surface,

\[
E=\frac12mv_0^2-\frac{GMm}{R}
\]

Conservation of mechanical energy gives

\[
\frac12mv_0^2-\frac{GMm}{R}
=
-\frac{GMm}{5R}
\]

Divide by \(m\) and rearrange:

\[
\frac12v_0^2
=
\frac{GM}{R}-\frac{GM}{5R}
\]

\[
\frac12v_0^2
=
\frac{4GM}{5R}
\]

Therefore,

\[
v_0^2=\frac{8GM}{5R}
\]

Step 3: Compare the launch speed with escape speed.

The surface escape speed satisfies

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

Taking the ratio,

\[
\frac{v_0^2}{v_{\mathrm{esc}}^2}
=
\frac{8/5}{2}
\]

\[
\frac{v_0^2}{v_{\mathrm{esc}}^2}
=\frac45
\]

Taking the square root:

\[
\frac{v_0}{v_{\mathrm{esc}}}
=\sqrt{0.8}
\]

\[
v_0\approx0.894v_{\mathrm{esc}}
\]

The probe was launched below escape speed.

Step 4: Calculate the gravitational field strength at \(5R\).

Using the inverse-square relationship,

\[
\frac{g_{5R}}{g_s}
=
\frac{R^2}{(5R)^2}
\]

\[
\frac{g_{5R}}{g_s}=\frac1{25}
\]

The field strength is still non-zero.

The tempting conclusion is that travelling such a large distance means the probe must have escaped. However, distance travelled is not the deciding factor. Its total mechanical energy is negative, so it must reach a finite turning point and return.

Question 3

A planet has a radius of \(6.0\times10^6\ \text{m}\) and a surface gravitational field strength of \(9.0\ \text{m s}^{-2}\).

A \(200\ \text{kg}\) instrument is moved slowly from the planet’s surface to a distance of \(3R\) from its centre.

Calculate:

a. The change in gravitational potential energy.

b. The work done by gravity.

c. The gravitational field strength at \(3R\).

A student argues that because gravitational field strength decreases to one-ninth of its surface value, gravitational potential energy must also decrease to one-ninth of its initial magnitude.

Explain why this reasoning is incorrect.

Solution 3

The instrument gains \(7.20\times10^9\ \text{J}\) of gravitational potential energy. Gravity does \(-7.20\times10^9\ \text{J}\) of work, and the gravitational field strength at \(3R\) is \(1.0\ \text{m s}^{-2}\). The student’s mistake is treating potential energy and field strength as though they have the same dependence on distance.

Step 1: Calculate \(GM/R\).

We know that surface gravitational field strength is

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

Rearrange:

\[
\frac{GM}{R}=g_sR
\]

Substitute:

\[
\frac{GM}{R}
=(9.0)(6.0\times10^6)
\]

\[
\frac{GM}{R}=5.4\times10^7\ \text{J kg}^{-1}
\]

Step 2: Calculate the change in gravitational potential energy.

Use

\[
\Delta U=U_f-U_i
\]

The initial position is \(R\), and the final position is \(3R\).

Therefore,

\[
\Delta U
=
-\frac{GMm}{3R}
-\left(-\frac{GMm}{R}\right)
\]

Simplify:

\[
\Delta U
=
\frac{2GMm}{3R}
\]

Substitute:

\[
\Delta U
=
\frac23(5.4\times10^7)(200)
\]

\[
\boxed{\Delta U=7.20\times10^9\ \text{J}}
\]

The positive result means the instrument gains gravitational potential energy as it moves away from the planet.

Step 3: Calculate the work done by gravity.

For a conservative gravitational field,

\[
W_g=-\Delta U
\]

Therefore,

\[
W_g=-7.20\times10^9\ \text{J}
\]

Gravity does negative work because the displacement is away from the planet while the gravitational force points towards it.

Since the instrument moves slowly, an external agent must supply approximately \(+7.20\times10^9\ \text{J}\) of work.

Step 4: Calculate gravitational field strength at \(3R\).

Use

\[
g_{3R}=\frac{GM}{(3R)^2}
\]

Therefore,

\[
g_{3R}=\frac{g_s}{9}
\]

\[
g_{3R}=\frac{9.0}{9}
\]

\[
\boxed{g_{3R}=1.0\ \text{m s}^{-2}}
\]

Step 5: Identify the misconception.

The student is correct that gravitational field strength decreases to one-ninth of its original value.

However, gravitational potential energy varies with \(1/r\), not \(1/r^2\).

At \(3R\), the potential energy is

\[
U_{3R}=-\frac{GMm}{3R}
\]

Its magnitude is one-third of its original value, not one-ninth.

The difference matters because gravitational potential energy measures the accumulated energy change between positions, whereas field strength describes the gravitational effect at one particular position.

Question 4

A \(900\ \text{kg}\) satellite moves in a circular orbit at a radius of \(2R\) around a planet.

For this planet,

\[
\frac{GM}{R}=4.00\times10^7\ \text{J kg}^{-1}
\]

Calculate:

a. The satellite’s kinetic energy.

b. Its gravitational potential energy.

c. Its total mechanical energy.

Explain why the negative total mechanical energy does not mean that the satellite must fall directly towards the planet’s surface.

Solution 4

The satellite has kinetic energy of \(+9.00\times10^9\ \text{J}\), gravitational potential energy of \(-1.80\times10^{10}\ \text{J}\), and total mechanical energy of \(-9.00\times10^9\ \text{J}\). It remains in a circular orbit because its tangential velocity and gravitational acceleration continually change the direction of its motion.

Step 1: Determine the satellite’s orbital speed.

For circular motion, gravitational force provides the required centripetal force:

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

Cancel \(m\) and rearrange:

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

At \(r=2R\),

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

Step 2: Calculate kinetic energy.

\[
K=\frac12mv^2
\]

Substitute the orbital speed expression:

\[
K=\frac{GMm}{4R}
\]

Using the given information:

\[
K=\frac{(4.00\times10^7)(900)}{4}
\]

\[
\boxed{K=9.00\times10^9\ \text{J}}
\]

Step 3: Calculate gravitational potential energy.

Use

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

At \(r=2R\),

\[
U=-\frac{GMm}{2R}
\]

Substitute:

\[
U=-\frac{(4.00\times10^7)(900)}{2}
\]

\[
\boxed{U=-1.80\times10^{10}\ \text{J}}
\]

Step 4: Calculate total mechanical energy.

\[
E=K+U
\]

\[
E=9.00\times10^9-1.80\times10^{10}
\]

\[
\boxed{E=-9.00\times10^9\ \text{J}}
\]

The satellite has negative total mechanical energy, meaning it is gravitationally bound.

However, bound does not mean falling directly into the planet.

The satellite has substantial sideways velocity. Gravity continuously accelerates it towards the planet, changing the direction of its velocity and keeping it in circular motion.

This is why the sign of total mechanical energy tells us whether motion is bound, not whether the object is moving directly towards the planet.

Question 5

An examiner introduces a different convention for measuring gravitational potential energy.

Instead of defining gravitational potential energy to be zero at infinity, the examiner chooses

\[
U_B(2R)=0
\]

For this convention, the gravitational potential energy is

\[
U_B(r)=-\frac{GMm}{r}+\frac{GMm}{2R}
\]

A spacecraft is launched radially outwards from the planet’s surface at \(0.80\) times the usual surface escape speed.

A student calculates that its total mechanical energy is positive using the examiner’s convention and concludes that the spacecraft must escape.

Determine whether the student is correct.

Your answer must:

a. Calculate the total mechanical energy using the usual convention \(U(\infty)=0\).

b. Calculate the total mechanical energy using the examiner’s convention.

c. Determine the gravitational potential energy at infinity using the examiner’s convention.

d. Calculate the maximum distance reached by the spacecraft.

e. Explain why the sign of total mechanical energy alone cannot determine escape unless the reference point is specified.

For this question, assume an ideal spherical planet, radial motion, and conservation of mechanical energy.

Solution 5

The student is incorrect. The spacecraft does not escape, despite having positive total mechanical energy under the examiner’s convention. It reaches a maximum distance of approximately \(2.78R\).

The crucial issue is that the condition \(E>0\) indicates escape only when gravitational potential energy has been defined as zero at infinity.

Step 1: Calculate the spacecraft’s initial kinetic energy.

Let the usual surface escape speed be \(v_{\mathrm{esc}}\).

The initial speed is

\[
v_0=0.80v_{\mathrm{esc}}
\]

We know

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

Therefore,

\[
v_0^2=0.64\frac{2GM}{R}
\]

The initial kinetic energy is

\[
K_0=\frac12mv_0^2
\]

Substituting:

\[
K_0=0.64\frac{GMm}{R}
\]

Step 2: Calculate total mechanical energy using the usual convention.

With gravitational potential energy zero at infinity,

\[
U_A(R)=-\frac{GMm}{R}
\]

Therefore,

\[
E_A=K_0+U_A
\]

\[
E_A=
0.64\frac{GMm}{R}
-\frac{GMm}{R}
\]

\[
\boxed{E_A=-0.36\frac{GMm}{R}}
\]

The total mechanical energy is negative, so the spacecraft is bound.

Step 3: Calculate total mechanical energy using the examiner’s convention.

The alternative potential energy is

\[
U_B(r)=-\frac{GMm}{r}+\frac{GMm}{2R}
\]

At the planet’s surface, \(r=R\):

\[
U_B(R)
=
-\frac{GMm}{R}
+\frac{GMm}{2R}
\]

Therefore,

\[
U_B(R)=-0.50\frac{GMm}{R}
\]

The total mechanical energy becomes

\[
E_B=K_0+U_B(R)
\]

\[
E_B=
0.64\frac{GMm}{R}
-0.50\frac{GMm}{R}
\]

\[
\boxed{E_B=+0.14\frac{GMm}{R}}
\]

The examiner’s convention gives a positive total mechanical energy.

This is where the student’s reasoning fails.

Step 4: Calculate the gravitational potential energy at infinity under the new convention.

As \(r\) approaches infinity,

\[
-\frac{GMm}{r}\rightarrow0
\]

Therefore,

\[
U_B(\infty)=\frac{GMm}{2R}
\]

This is not zero.

For the spacecraft to reach infinity, its kinetic energy there must be at least zero.

Consequently, the correct escape condition under this convention is

\[
E_B\geq U_B(\infty)
\]

However,

\[
0.14\frac{GMm}{R}
<
0.50\frac{GMm}{R}
\]

The spacecraft does not have enough energy to reach infinity.

Step 5: Calculate its maximum distance.

At the maximum distance, the spacecraft momentarily stops, so

\[
K=0
\]

Therefore,

\[
E_B=U_B(r_{\max})
\]

Substitute:

\[
0.14\frac{GMm}{R}
=
-\frac{GMm}{r_{\max}}
+\frac{GMm}{2R}
\]

Divide by \(GMm/R\):

\[
0.14=-\frac{R}{r_{\max}}+0.50
\]

Rearrange:

\[
\frac{R}{r_{\max}}=0.36
\]

Therefore,

\[
r_{\max}=\frac{R}{0.36}
\]

\[
\boxed{r_{\max}=2.78R}
\]

The spacecraft reaches a finite turning point and returns.

Step 6: Explain what the change of reference actually did.

The examiner added the same constant to gravitational potential energy at every position.

This changes the numerical value of total mechanical energy, but it does not change the physical energy difference between two positions.

It also does not change the gravitational force. The force depends on how potential energy changes with position, not on the arbitrary constant added to every value.

This is why the usual convention is so useful:

\[
U(\infty)=0
\]

With this reference, the sign of total mechanical energy gives us a direct test:

• \(E<0\): bound.
• \(E=0\): minimum-energy escape.
• \(E>0\): unbound.

The deeper lesson is that zero total mechanical energy and zero gravitational field strength are entirely different ideas. The first describes the balance between kinetic and potential energy. The second describes the gravitational influence at a location.

Once you can distinguish these, the next useful step is calculating how gravitational work changes an object’s energy as it moves between two orbital radii. See How to Calculate Gravitational Work Between Two Orbital Radii.