How to Read an Escape-Energy Graph with a Negative Potential Axis

A spacecraft is moving away from a planet. Its gravitational potential energy is \(-4.0\ \text{GJ}\), but it also has \(1.8\ \text{GJ}\) of kinetic energy. Is its total energy negative? Can it escape?

The answer is not obvious if you're looking only at the negative numbers on an energy graph. The key is to distinguish **gravitational potential energy from total mechanical energy**, then read the vertical gap between the two as kinetic energy.

## How to read an escape-energy graph

Start with the graph's two axes:

- The horizontal axis represents distance \(r\) from the centre of the planet, measured in metres.
- The vertical axis represents energy, measured in joules.

The gravitational potential energy curve sits below zero and approaches zero as the distance increases towards infinity. A horizontal line represents the object's total mechanical energy.

The most useful relationship is:

\[
E=K+U
\]

Rearranging:

\[
\boxed{K=E-U}
\]

Here:

- \(E\) is total mechanical energy, in joules (J).
- \(K\) is kinetic energy, in joules (J).
- \(U\) is gravitational potential energy, in joules (J).

**The vertical gap between the total-energy line and the potential-energy curve is the object's kinetic energy.**

This is the central rule for reading the graph.

[IMAGE: Annotated gravitational escape-energy graph with distance from the planet's centre on the horizontal axis and energy on the vertical axis. Show the negative gravitational potential energy curve U=-GMm/r rising towards zero from below, the zero-energy axis, and a horizontal negative total-energy line E. At a sample radius, label the vertical gap between U and E as positive kinetic energy K=E-U. Clearly distinguish the potential curve, total-energy line, and zero reference.]

### Why is the potential-energy curve negative?

We choose gravitational potential energy to be zero at infinite separation:

\[
U(\infty)=0
\]

With this convention, the gravitational potential energy of an object of mass \(m\), at a distance \(r\) from the centre of a spherical planet of mass \(M\), is:

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

where \(G\) is the universal gravitational constant.

The negative sign tells us that the object is gravitationally bound in the sense that energy must be supplied to move it from its current position to infinity without leaving any kinetic energy behind.

As the object moves further away, \(U\) increases towards zero. For example:

| Distance from planet's centre | Gravitational potential energy |
|---|---:|
| \(r\) | \(-4.0\ \text{GJ}\) |
| \(2r\) | \(-2.0\ \text{GJ}\) |
| \(4r\) | \(-1.0\ \text{GJ}\) |
| Infinity | \(0\ \text{GJ}\) |

Notice that the potential energy becomes less negative as the object moves away. It is increasing, even though its numerical value remains negative.

This is why the negative vertical axis matters. A point higher on the graph represents greater energy, even if both values are below zero.

[IMAGE: Four-position illustration of a spacecraft moving progressively farther from a spherical planet. Show the centre-to-spacecraft distance increasing, with corresponding potential energy values -4.0 GJ, -2.0 GJ, -1.0 GJ, and approaching 0 GJ. Emphasise the increasing distance and the potential energy becoming less negative.]

If the negative sign still feels counterintuitive, see [Why Gravitational Potential Energy Is Negative Around a Planet](https://tutorgum.com/learn/resources/why-gravitational-potential-energy-is-negative-around-a-planet/). The important point here is how that negative curve interacts with the object's total energy.

### The three energy conditions you need to recognise

The total mechanical energy determines whether an object can escape in an ideal gravitational field.

| Total mechanical energy | What the graph tells you | Outcome |
|---|---|---|
| \(E<0\) | Total-energy line is below zero | Gravitationally bound |
| \(E=0\) | Total-energy line lies on zero | Just enough energy to escape |
| \(E>0\) | Total-energy line is above zero | Escapes with kinetic energy remaining |

Why does this work?

At infinity, gravitational potential energy approaches zero. Therefore, conservation of mechanical energy gives:

\[
E=K_\infty+0
\]

So:

\[
K_\infty=E
\]

This means:

- If \(E<0\), the object would need negative kinetic energy to reach infinity. That is impossible, so it cannot escape.
- If \(E=0\), it reaches infinity with its speed approaching zero. This is the minimum-energy escape condition.
- If \(E>0\), it can reach infinity with positive kinetic energy remaining.

The important distinction is that **negative gravitational potential energy does not automatically mean negative total energy**.

An object can have negative potential energy and still escape if its positive kinetic energy is large enough.

## A reliable method for reading the graph

When you are given an escape-energy graph in an HSC Physics question, use this sequence.

**Step 1: Identify the zero-energy reference.**

For this type of graph, gravitational potential energy is defined as zero at infinity. The horizontal axis at zero energy is therefore your escape threshold.

**Step 2: Identify the potential-energy curve.**

The gravitational potential energy curve is negative and approaches zero as \(r\) increases.

**Step 3: Find the total-energy line.**

This is the horizontal line representing the object's conserved total mechanical energy, assuming no external work or energy losses.

**Step 4: Find the kinetic energy.**

At the radius you're interested in, calculate the vertical separation:

\[
K=E-U
\]

Do not add the magnitudes of the two negative values. Pay attention to their signs.

**Step 5: Determine whether escape is possible.**

Compare the total-energy line with zero, not the potential-energy curve with zero.

**Step 6: Look for an intersection if the motion is radial.**

Where \(E=U\), kinetic energy is zero. For an object moving directly away from the planet with no sideways velocity, this is its maximum distance before it stops momentarily and falls back.

This last step has an important assumption: a non-radial object can retain sideways kinetic energy at its maximum distance. We will return to that distinction later.

## Worked example: A spacecraft with negative total energy

A \(100\ \text{kg}\) spacecraft is launched directly away from a planet. The planet's gravitational parameter is:

\[
GM=4.0\times10^{14}\ \text{m}^3\text{s}^{-2}
\]

At the instant considered, the spacecraft is \(1.0\times10^7\ \text{m}\) from the planet's centre and travelling outwards at \(6.0\times10^3\ \text{m s}^{-1}\).

Determine:

a. its gravitational potential energy  
b. its kinetic energy  
c. its total mechanical energy  
d. whether it can escape.

**Step 1**: Calculate gravitational potential energy.

Use:

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

Substitute:

\[
\begin{aligned}
U&=-\frac{(4.0\times10^{14})(100)}
{1.0\times10^7}\\
&=-4.0\times10^9\ \text{J}
\end{aligned}
\]

The spacecraft has \(-4.0\ \text{GJ}\) of gravitational potential energy.

**Step 2**: Calculate kinetic energy.

Use:

\[
K=\frac12mv^2
\]

\[
\begin{aligned}
K&=\frac12(100)(6.0\times10^3)^2\\
&=50(36\times10^6)\\
&=1.8\times10^9\ \text{J}
\end{aligned}
\]

Its kinetic energy is \(+1.8\ \text{GJ}\).

**Step 3**: Calculate total mechanical energy.

Add the signed energy values:

\[
\begin{aligned}
E&=K+U\\
&=1.8\times10^9-4.0\times10^9\\
&=-2.2\times10^9\ \text{J}
\end{aligned}
\]

The total mechanical energy is \(-2.2\ \text{GJ}\).

**Step 4**: Determine whether the spacecraft escapes.

Since:

\[
E<0
\]

the spacecraft is gravitationally bound. It does not have enough mechanical energy to reach infinity.

Notice the graph interpretation:

- The potential-energy curve is at \(-4.0\ \text{GJ}\).
- The total-energy line is at \(-2.2\ \text{GJ}\).
- The vertical gap is \(1.8\ \text{GJ}\), which is the spacecraft's positive kinetic energy.

Both energy values are negative, but the spacecraft is still moving because its total energy is greater than its potential energy.

[IMAGE: Worked-example energy graph for the 100 kg spacecraft. Show the potential curve at -4.0 GJ and the horizontal total-energy line at -2.2 GJ at r=1.0 x 10^7 m. Annotate the 1.8 GJ vertical separation as kinetic energy. Show the zero-energy line above both.]

## What happens at the turning point?

The spacecraft in the previous example cannot escape. But how far can it travel?

For this calculation, assume it continues moving directly away from the planet, with no sideways velocity, and that mechanical energy is conserved.

At its maximum distance, it momentarily stops. Therefore:

\[
K=0
\]

Since:

\[
E=K+U
\]

at this instant:

\[
E=U
\]

Substitute the gravitational potential energy equation:

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

Rearrange:

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

Using the previous example:

\[
\begin{aligned}
r_{\max}
&=-\frac{(4.0\times10^{14})(100)}
{-2.2\times10^9}\\
&=1.82\times10^7\ \text{m}
\end{aligned}
\]

The spacecraft reaches approximately \(1.82\times10^7\ \text{m}\) from the planet's centre before momentarily stopping.

On the graph, this is the intersection between the total-energy line and the gravitational potential-energy curve.

[IMAGE: Annotated radial escape-energy graph showing the negative total-energy line intersecting the gravitational potential curve at a finite maximum radius. Mark the launch position at 1.0 x 10^7 m and the turning point at 1.82 x 10^7 m. Show the positive kinetic-energy gap shrinking to zero at the intersection. Do not show the curve extending into physically inaccessible radii as though the spacecraft travels there.]

This is a useful result, but don't confuse the turning point with the escape condition.

The turning point occurs where the total energy equals the potential energy. Escape occurs when the total energy is at least zero.

### Worked example: A spacecraft escapes even though its potential energy is negative

A \(100\ \text{kg}\) spacecraft is now travelling at \(1.0\times10^4\ \text{m s}^{-1}\) at the same distance from the same planet.

The planet has \(GM=4.0\times10^{14}\ \text{m}^3\text{s}^{-2}\), and the spacecraft is \(1.0\times10^7\ \text{m}\) from its centre.

Determine its total mechanical energy, whether it can escape, and its speed at infinity.

**Step 1**: Calculate gravitational potential energy.

The position and planet have not changed, so:

\[
U=-4.0\times10^9\ \text{J}
\]

**Step 2**: Calculate kinetic energy.

\[
\begin{aligned}
K&=\frac12mv^2\\
&=\frac12(100)(1.0\times10^4)^2\\
&=5.0\times10^9\ \text{J}
\end{aligned}
\]

**Step 3**: Calculate total mechanical energy.

\[
\begin{aligned}
E&=K+U\\
&=5.0\times10^9-4.0\times10^9\\
&=+1.0\times10^9\ \text{J}
\end{aligned}
\]

The total mechanical energy is positive, so the spacecraft can escape.

**Step 4**: Calculate its speed at infinity.

At infinity, gravitational potential energy approaches zero. Therefore:

\[
K_\infty=E=1.0\times10^9\ \text{J}
\]

Using kinetic energy:

\[
\begin{aligned}
K_\infty&=\frac12mv_\infty^2\\
1.0\times10^9&=\frac12(100)v_\infty^2\\
v_\infty^2&=2.0\times10^7\\
v_\infty&=4.47\times10^3\ \text{m s}^{-1}
\end{aligned}
\]

The spacecraft escapes with a speed approaching \(4.47\ \text{km s}^{-1}\) at infinity.

Notice the difference between this example and the previous one:

- Both spacecraft start with negative gravitational potential energy.
- The first has negative total energy and cannot escape.
- The second has positive total energy and escapes.

The potential-energy curve has not changed. Only the total-energy line has changed.

[IMAGE: Comparison of two total-energy lines on the same negative gravitational potential energy curve. One horizontal line is at -2.2 GJ and the other at +1.0 GJ. Show the zero-energy axis clearly. The negative line intersects the potential curve at a finite radius, while the positive line remains above the entire negative potential curve. Label the two lines as bound and escaping without implying that the positive line is a potential-energy curve.]

## What changes when the total energy is zero?

The boundary between bound and escaping motion is particularly important.

Consider an object with exactly zero total mechanical energy:

\[
E=0
\]

At any finite distance from the planet, its gravitational potential energy is negative.

Therefore:

\[
K=E-U
\]

\[
K=0-U=-U
\]

Since \(U\) is negative, \(K\) is positive.

The object is still moving at every finite distance. Its kinetic energy gradually decreases as it travels outwards.

Only as its distance approaches infinity does its kinetic energy approach zero.

This is the minimum-energy escape condition. It is also the condition used to derive escape speed.

[IMAGE: Negative gravitational potential energy curve with the total-energy line exactly on the zero-energy axis. Show a positive kinetic-energy gap at a finite radius that gradually narrows as the curve approaches zero at infinity. Label the asymptotic escape condition E=0.]

A common mistake is to think that an object with \(E=0\) must have stopped moving already.

It hasn't. Zero total energy is not zero kinetic energy at every position. It is the limiting condition at infinity.

For a derivation connecting this graph to the escape-speed equation, 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/).

## Common mistakes when reading a negative energy axis

| Tempting interpretation | Correct interpretation |
|---|---|
| Negative potential energy means the object cannot escape. | Escape depends on total mechanical energy, not potential energy alone. |
| A higher negative value represents less energy. | \(-2\ \text{J}\) is greater than \(-8\ \text{J}\). |
| The vertical distance between zero and the potential curve is kinetic energy. | Kinetic energy is the gap between total energy and potential energy. |
| An object with negative total energy is stationary. | It can be moving rapidly while remaining gravitationally bound. |
| Escape happens when potential energy reaches zero. | Escape is determined by total energy. Potential energy approaches zero at infinity. |
| The intersection of the total-energy line and potential curve is always the maximum radius. | This is true for radial motion with no sideways velocity, but not generally for orbital motion. |
| The horizontal total-energy line always stays in the same position. | External work, such as an engine firing, can change total mechanical energy. |

## Assumptions behind the graph

The calculations and interpretations in this article use an ideal gravitational model.

- The planet is treated as a spherical body with mass \(M\).
- The escaping object has mass \(m\).
- Distance \(r\) is measured from the planet's centre, not its surface.
- Gravitational potential energy is zero at infinity.
- The gravitational field is treated using Newtonian gravity.
- Mechanical energy is conserved unless external work is explicitly introduced.
- Atmospheric drag, collisions, propulsion, and other gravitational bodies are ignored.

One further distinction is worth remembering: negative total energy means gravitationally bound, not necessarily moving directly back towards the planet. A spacecraft can have negative total energy and remain in an elliptical or circular orbit.

The simple potential-energy graph describes the gravitational potential energy as a function of radius. Interpreting its intersections as actual radial turning points requires the additional assumption of purely radial motion.

That distinction becomes important in the final practice question.

## Questions and solutions

The following questions use illustrative data for hypothetical planets. Assume gravitational potential energy is zero at infinity throughout.

### Question 1

An object is at a position where its gravitational potential energy is \(-9.0\ \text{MJ}\). Its total mechanical energy is \(-3.0\ \text{MJ}\).

Calculate its kinetic energy and determine whether the object can escape.

[IMAGE: Simple negative gravitational potential energy graph showing a potential curve at -9.0 MJ and a horizontal total-energy line at -3.0 MJ at a marked radius. Include the zero-energy axis. Do not label the kinetic-energy gap or state the escape outcome.]

### Solution 1

The object's kinetic energy is \(+6.0\ \text{MJ}\), and it is gravitationally bound.

Start with conservation of mechanical energy:

\[
E=K+U
\]

Rearrange:

\[
K=E-U
\]

Substitute the signed values:

\[
\begin{aligned}
K&=-3.0-(-9.0)\\
&=+6.0\ \text{MJ}
\end{aligned}
\]

The positive kinetic energy confirms that the object is moving.

However, its total mechanical energy is negative:

\[
E=-3.0\ \text{MJ}<0
\]

Therefore, it cannot escape under the stated assumptions.

The tempting mistake is to see two negative energy values and assume the kinetic energy must also be negative. The subtraction of the negative potential energy produces a positive kinetic energy.

### Question 2

A \(200\ \text{kg}\) probe is launched directly away from a planet.

The planet has a gravitational parameter of:

\[
GM=3.2\times10^{14}\ \text{m}^3\text{s}^{-2}
\]

At one instant, the probe is \(8.0\times10^6\ \text{m}\) from the planet's centre and travelling outwards at \(7.0\times10^3\ \text{m s}^{-1}\).

Calculate:

a. the probe's gravitational potential energy  
b. its kinetic energy  
c. its total mechanical energy  
d. its maximum distance from the planet's centre.

[IMAGE: Physical setup of a spherical planet and a probe moving directly outwards. Label the probe mass as 200 kg, its centre-to-centre distance as 8.0 x 10^6 m, and its outward speed as 7.0 x 10^3 m/s. Do not show a calculated turning point.]

### Solution 2

The probe has a total mechanical energy of \(-3.1\times10^9\ \text{J}\), so it is bound. Its maximum distance from the planet's centre is approximately \(2.1\times10^7\ \text{m}\).

**Step 1**: Calculate gravitational potential energy.

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

\[
\begin{aligned}
U&=-\frac{(3.2\times10^{14})(200)}
{8.0\times10^6}\\
&=-8.0\times10^9\ \text{J}
\end{aligned}
\]

**Step 2**: Calculate kinetic energy.

\[
K=\frac12mv^2
\]

\[
\begin{aligned}
K&=\frac12(200)(7.0\times10^3)^2\\
&=4.9\times10^9\ \text{J}
\end{aligned}
\]

**Step 3**: Calculate total mechanical energy.

\[
\begin{aligned}
E&=K+U\\
&=4.9\times10^9-8.0\times10^9\\
&=-3.1\times10^9\ \text{J}
\end{aligned}
\]

Since \(E<0\), the probe cannot escape.

**Step 4**: Find the maximum distance.

At the maximum distance, the probe momentarily stops, so:

\[
K=0
\]

Therefore:

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

Rearranging:

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

Substitute:

\[
\begin{aligned}
r_{\max}
&=-\frac{(3.2\times10^{14})(200)}
{-3.1\times10^9}\\
&=2.06\times10^7\ \text{m}
\end{aligned}
\]

The probe reaches approximately \(2.1\times10^7\ \text{m}\) from the planet's centre.

The important graph-reading step is recognising that the maximum distance occurs where the negative total-energy line meets the potential-energy curve. The object has used all its kinetic energy by that point.

### Question 3

A \(50\ \text{kg}\) probe is moving away from a planet with:

\[
GM=5.0\times10^{14}\ \text{m}^3\text{s}^{-2}
\]

At a distance of \(1.0\times10^7\ \text{m}\) from the planet's centre, its speed is \(1.2\times10^4\ \text{m s}^{-1}\).

Determine:

a. its total mechanical energy  
b. its kinetic energy when it reaches \(2.5\times10^7\ \text{m}\)  
c. its speed at that position  
d. its speed at infinity.

A student claims the probe cannot escape because its gravitational potential energy is still negative at \(2.5\times10^7\ \text{m}\).

Explain why this conclusion is incorrect.

[IMAGE: Spherical planet with a probe moving radially outwards. Mark the initial centre-to-probe distance as 1.0 x 10^7 m and a second position farther out at 2.5 x 10^7 m. Show outward motion and both positions without adding energy values or speeds.]

### Solution 3

The probe has positive total mechanical energy of \(+1.1\times10^9\ \text{J}\), so it escapes. Its speed at \(2.5\times10^7\ \text{m}\) is approximately \(9.2\times10^3\ \text{m s}^{-1}\), and its speed at infinity approaches \(6.6\times10^3\ \text{m s}^{-1}\).

**Step 1**: Calculate the initial gravitational potential energy.

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

\[
\begin{aligned}
U&=-\frac{(5.0\times10^{14})(50)}
{1.0\times10^7}\\
&=-2.5\times10^9\ \text{J}
\end{aligned}
\]

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

\[
\begin{aligned}
K&=\frac12mv^2\\
&=\frac12(50)(1.2\times10^4)^2\\
&=3.6\times10^9\ \text{J}
\end{aligned}
\]

**Step 3**: Calculate total mechanical energy.

\[
\begin{aligned}
E&=K+U\\
&=3.6\times10^9-2.5\times10^9\\
&=+1.1\times10^9\ \text{J}
\end{aligned}
\]

The positive total energy means the probe can escape.

**Step 4**: Calculate potential energy at the second position.

\[
\begin{aligned}
U_2&=-\frac{(5.0\times10^{14})(50)}
{2.5\times10^7}\\
&=-1.0\times10^9\ \text{J}
\end{aligned}
\]

**Step 5**: Calculate kinetic energy at the second position.

Total mechanical energy remains constant:

\[
E=K_2+U_2
\]

Therefore:

\[
\begin{aligned}
K_2&=E-U_2\\
&=1.1\times10^9-(-1.0\times10^9)\\
&=2.1\times10^9\ \text{J}
\end{aligned}
\]

**Step 6**: Calculate speed at the second position.

\[
K_2=\frac12mv_2^2
\]

\[
\begin{aligned}
v_2&=\sqrt{\frac{2K_2}{m}}\\
&=\sqrt{\frac{2(2.1\times10^9)}{50}}\\
&=9.17\times10^3\ \text{m s}^{-1}
\end{aligned}
\]

**Step 7**: Calculate speed at infinity.

At infinity:

\[
U_\infty=0
\]

Therefore:

\[
K_\infty=E=1.1\times10^9\ \text{J}
\]

Using kinetic energy:

\[
\begin{aligned}
v_\infty&=\sqrt{\frac{2E}{m}}\\
&=\sqrt{\frac{2(1.1\times10^9)}{50}}\\
&=6.63\times10^3\ \text{m s}^{-1}
\end{aligned}
\]

The student's mistake is confusing negative potential energy with negative total energy.

At the second position, the probe still has negative gravitational potential energy, but its positive kinetic energy is larger. Its total energy remains positive, so escape is possible.

### Question 4

A \(250\ \text{kg}\) spacecraft is moving directly away from a planet.

The planet has:

\[
GM=4.0\times10^{14}\ \text{m}^3\text{s}^{-2}
\]

The spacecraft is initially \(1.0\times10^7\ \text{m}\) from the planet's centre and travelling at \(8.0\times10^3\ \text{m s}^{-1}\).

A thruster then performs \(3.0\times10^9\ \text{J}\) of positive work on the spacecraft. Assume its mass does not change.

Determine:

a. its total mechanical energy before the thruster fires  
b. the minimum additional work required for it to escape  
c. its speed at infinity after the actual thruster burn.

Explain how the total-energy line changes on the graph.

[IMAGE: Spacecraft at a fixed radial position from a spherical planet with its engine firing briefly. Show the planet, spacecraft, centre-to-centre distance of 1.0 x 10^7 m, and direction of the engine thrust. Do not show energy values or the final trajectory.]

### Solution 4

The spacecraft initially has total mechanical energy of \(-2.0\times10^9\ \text{J}\). It needs at least \(2.0\times10^9\ \text{J}\) of additional work to escape. After receiving \(3.0\times10^9\ \text{J}\), it escapes with a speed approaching \(2.8\times10^3\ \text{m s}^{-1}\) at infinity.

**Step 1**: Calculate initial gravitational potential energy.

\[
\begin{aligned}
U&=-\frac{(4.0\times10^{14})(250)}
{1.0\times10^7}\\
&=-1.0\times10^{10}\ \text{J}
\end{aligned}
\]

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

\[
\begin{aligned}
K&=\frac12(250)(8.0\times10^3)^2\\
&=8.0\times10^9\ \text{J}
\end{aligned}
\]

**Step 3**: Calculate initial total mechanical energy.

\[
\begin{aligned}
E_i&=K+U\\
&=8.0\times10^9-1.0\times10^{10}\\
&=-2.0\times10^9\ \text{J}
\end{aligned}
\]

The spacecraft is initially bound.

**Step 4**: Calculate the minimum additional work required.

For the spacecraft to just escape:

\[
E_f=0
\]

Therefore, the minimum energy increase is:

\[
\begin{aligned}
W_{\min}&=E_f-E_i\\
&=0-(-2.0\times10^9)\\
&=2.0\times10^9\ \text{J}
\end{aligned}
\]

The thruster must supply at least \(2.0\ \text{GJ}\) of work for escape.

**Step 5**: Calculate the new total mechanical energy.

The actual thruster supplies \(3.0\times10^9\ \text{J}\).

Therefore:

\[
\begin{aligned}
E_f&=E_i+W\\
&=-2.0\times10^9+3.0\times10^9\\
&=+1.0\times10^9\ \text{J}
\end{aligned}
\]

The spacecraft now has positive total mechanical energy.

**Step 6**: Calculate its speed at infinity.

At infinity:

\[
K_\infty=E_f
\]

So:

\[
\begin{aligned}
\frac12mv_\infty^2&=1.0\times10^9\\
v_\infty&=\sqrt{\frac{2(1.0\times10^9)}{250}}\\
&=2.83\times10^3\ \text{m s}^{-1}
\end{aligned}
\]

The spacecraft escapes with a speed approaching \(2.83\ \text{km s}^{-1}\).

**What happens on the graph?**

The gravitational potential-energy curve does not move. The planet's mass, spacecraft's mass, and position-dependent gravitational interaction have not changed.

Instead, the total-energy line moves upwards:

- Before the burn, \(E=-2.0\ \text{GJ}\).
- After the burn, \(E=+1.0\ \text{GJ}\).

The thruster has supplied enough energy to move the spacecraft from the bound region into the escaping region.

[IMAGE: Annotated before-and-after energy graph for Question 4. Use one unchanged negative gravitational potential energy curve. Show the initial total-energy line at -2.0 GJ and the final line at +1.0 GJ, with an upward arrow labelled +3.0 GJ work. Clearly show the zero-energy escape boundary.]

The tempting assumption is that the potential-energy curve must change when the spacecraft receives energy. It doesn't. The energy supplied changes the spacecraft's total mechanical energy, not the gravitational potential-energy function.

### Question 5

Two identical \(100\ \text{kg}\) spacecraft are at the same distance from a planet:

\[
r=1.50\times10^7\ \text{m}
\]

The planet has:

\[
GM=4.0\times10^{14}\ \text{m}^3\text{s}^{-2}
\]

Both spacecraft have a speed of \(3.06\times10^3\ \text{m s}^{-1}\).

However, their directions of motion differ:

- Spacecraft A is travelling directly away from the planet.
- Spacecraft B is travelling perpendicular to the radius joining it to the planet. Its velocity is entirely tangential.

Assume no propulsion, drag, or other gravitational influences.

A student draws the gravitational potential-energy curve and a single total-energy line. They calculate where the two lines intersect and claim that both spacecraft will reach that same maximum distance.

Evaluate this claim.

Your answer should:

a. calculate the total mechanical energy of each spacecraft  
b. calculate the maximum radius predicted for spacecraft A  
c. determine whether spacecraft B reaches that radius  
d. explain why the same total-energy line does not guarantee the same radial motion.

[IMAGE: Question setup showing a spherical planet and two identical spacecraft at the same radial distance of 1.50 x 10^7 m. Spacecraft A has an outward radial velocity arrow. Spacecraft B has a perpendicular tangential velocity arrow. Both arrows should have the same labelled speed of 3.06 x 10^3 m/s. Do not draw either trajectory, orbital path, turning point, or answer.]

### Solution 5

The student's claim is incorrect. Spacecraft A reaches approximately \(1.82\times10^7\ \text{m}\), but spacecraft B does not. Spacecraft B is already at its maximum radius of \(1.50\times10^7\ \text{m}\).

This is a subtle limitation of reading a gravitational potential-energy graph: total mechanical energy alone does not determine the radial motion when angular momentum is involved.

**Step 1**: Calculate gravitational potential energy for both spacecraft.

They have identical masses and positions, so:

\[
\begin{aligned}
U&=-\frac{GMm}{r}\\
&=-\frac{(4.0\times10^{14})(100)}
{1.50\times10^7}\\
&=-2.67\times10^9\ \text{J}
\end{aligned}
\]

Both spacecraft have the same gravitational potential energy.

**Step 2**: Calculate kinetic energy.

Their speeds and masses are also identical:

\[
\begin{aligned}
K&=\frac12mv^2\\
&=\frac12(100)(3.06\times10^3)^2\\
&=4.68\times10^8\ \text{J}
\end{aligned}
\]

**Step 3**: Calculate total mechanical energy.

\[
\begin{aligned}
E&=K+U\\
&=4.68\times10^8-2.67\times10^9\\
&=-2.20\times10^9\ \text{J}
\end{aligned}
\]

Both spacecraft have the same negative total mechanical energy.

Both are gravitationally bound.

**Step 4**: Calculate the maximum radius for spacecraft A.

Spacecraft A has purely radial outward motion.

At its maximum radius, all kinetic energy has been converted into gravitational potential energy:

\[
E=U
\]

Therefore:

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

Rearrange:

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

Substitute:

\[
\begin{aligned}
r_{\max}
&=-\frac{(4.0\times10^{14})(100)}
{-2.20\times10^9}\\
&=1.82\times10^7\ \text{m}
\end{aligned}
\]

Spacecraft A reaches approximately \(1.82\times10^7\ \text{m}\).

**Step 5**: Examine the motion of spacecraft B.

Spacecraft B is travelling tangentially, so its velocity is perpendicular to the radius.

At that instant, its radial velocity is zero.

To determine what happens next, compare its tangential speed with the circular orbital speed at that radius:

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

Substitute:

\[
\begin{aligned}
v_{\text{circular}}
&=\sqrt{\frac{4.0\times10^{14}}
{1.50\times10^7}}\\
&=5.16\times10^3\ \text{m s}^{-1}
\end{aligned}
\]

But spacecraft B is travelling at only:

\[
v_B=3.06\times10^3\ \text{m s}^{-1}
\]

Its speed is below the circular orbital speed.

At this instant, gravity produces a greater inward acceleration than would be required for circular motion. The spacecraft therefore begins moving towards the planet.

Because its radial velocity was initially zero, and its radial acceleration is now inward, the starting position is its maximum radius. It is at the apocentre of its elliptical path.

Therefore, spacecraft B does not reach \(1.82\times10^7\ \text{m}\).

[IMAGE: Solution diagram comparing two trajectories from the same starting radius. Show spacecraft A travelling radially outwards before reaching its finite radial turning point at 1.82 x 10^7 m. Show spacecraft B at its initial outer turning point, then following an elliptical orbit inward. Label the tangential velocity and distinguish the two maximum radii. The central planet must be clearly shown.]

**Step 6**: Explain why the simple graph gives an incomplete prediction.

For both spacecraft, the total mechanical energy is:

\[
E=-2.20\times10^9\ \text{J}
\]

The gravitational potential-energy curve is also identical.

So why don't they reach the same maximum radius?

Because the equation:

\[
K=E-U
\]

describes the total kinetic energy, not just the radial component of kinetic energy.

For spacecraft A, all its kinetic energy is radial. At the maximum radius, its speed becomes zero.

For spacecraft B, angular momentum is conserved. It retains tangential motion as it travels around the planet. It cannot reach a finite radius where its total kinetic energy would have to become zero while its angular momentum remains non-zero.

The more complete model accounts for this additional constraint through an effective potential:

\[
U_{\text{eff}}=U+\frac{L^2}{2mr^2}
\]

Here, \(L\) is angular momentum. The additional positive term represents the constraint imposed by tangential motion.

This effective-potential treatment extends beyond the simple radial escape graph used in the basic method.

The crucial lesson is that **the total-energy line tells you whether motion is bound or unbound, but the gravitational potential-energy curve alone does not always tell you the maximum radius of a non-radial trajectory.**

## What to learn next

Once you can read the negative potential-energy axis, identify the total-energy line, and calculate the kinetic-energy gap, you have the foundation for escape-speed problems.

The next step is to connect the graph to the escape-speed equation:

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

This comes directly from setting total mechanical energy equal to zero at infinity.

For more practice connecting energy graphs, escape conditions, and launch speeds, see [How to Calculate Escape Speed from a Planet's Mass and Radius](https://tutorgum.com/learn/resources/hsc-physics-how-to-calculate-escape-speed-from-a-planet-s-mass-and-radius/).