Why mgh Fails at High Altitudes in HSC Physics

Suppose you lift a 1 kg instrument from Earth’s surface to an altitude equal to Earth’s radius. Using \(mgh\), with \(g=9.81\ \mathrm{m\,s^{-2}}\), you calculate an energy increase of approximately 62.5 million joules per kilogram. But there’s a problem: at that altitude, the instrument is twice as far from Earth’s centre, so gravitational acceleration has fallen to one-quarter of its surface value. The actual energy increase is only about 31.3 million joules per kilogram.

The formula \(mgh\) overestimates the energy required by a factor of two. The reason is that it assumes gravitational acceleration stays constant throughout the journey. That assumption works for small changes in altitude, but breaks down when the distance from Earth’s centre changes substantially.

01Why does \(mgh\) stop working at large altitudes?

Before looking at the equations, consider what happens when you lift an object in two different situations.

• Lifting a textbook: You raise it from the floor to a shelf. Its distance from Earth’s centre barely changes, so gravitational acceleration is effectively constant.
• Lifting a spacecraft: You move it thousands of kilometres above Earth’s surface. Its distance from Earth’s centre changes significantly, so the gravitational force becomes weaker as it travels.

The textbook and spacecraft experience the same underlying gravitational interaction. The difference is how much the gravitational field changes during the movement.

The assumption hidden inside \(mgh\)

The familiar gravitational potential energy equation is:

\[ \Delta U=mgh \]

Here:

• \(\Delta U\) is the change in gravitational potential energy, measured in joules (J).
• \(m\) is the object’s mass, in kilograms (kg).
• \(g\) is the gravitational acceleration, in metres per second squared (\(\mathrm{m\,s^{-2}}\)).
• \(h\) is the vertical displacement, in metres (m).

The equation assumes that \(g\) remains approximately constant over the displacement.

Near Earth’s surface, this is a very good approximation. Earth’s radius is approximately \(6.37\times10^{6}\ \mathrm{m}\), so lifting an object a few metres or even a few hundred metres barely changes its distance from Earth’s centre.

However, gravitational acceleration depends on distance from the centre of the attracting body:

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

Here, \(G\) is the universal gravitational constant, \(M\) is Earth’s mass, and \(r\) is the distance from Earth’s centre.

Notice the inverse-square relationship. If the distance from Earth’s centre doubles, gravitational acceleration becomes one-quarter of its original value.

This is the assumption that causes \(mgh\) to fail: it uses one value of \(g\) for an entire journey even though the actual gravitational field is changing.

02How Newton’s inverse-square model explains the problem

There is a useful piece of scientific reasoning behind this relationship.

Newton’s development of universal gravitation addressed a fundamental question: could the force that makes objects fall towards Earth also explain the Moon’s motion around Earth?

The Moon doesn’t travel in a straight line. It continually changes direction as it orbits Earth. This requires centripetal acceleration towards Earth’s centre.

If the same gravitational interaction acts on falling objects and the Moon, the gravitational acceleration should decrease with distance according to a consistent mathematical relationship.

The Moon test

Consider two sets of observations:

1. Near Earth’s surface, gravitational acceleration is approximately \(9.81\ \mathrm{m\,s^{-2}}\).
2. The Moon’s orbital radius is approximately 60 times Earth’s radius, and its orbital period is approximately 27.3 days.

The historical reasoning was to compare the acceleration predicted by an inverse-square gravitational field with the acceleration inferred from the Moon’s orbital motion.

If gravity follows an inverse-square relationship, the Moon’s gravitational acceleration should be approximately:

\[ g_{\text{Moon}}=\frac{g_{\text{surface}}}{60^{2}} \]

Therefore:

\[ g_{\text{Moon}}=\frac{9.81}{3600} \]

\[ g_{\text{Moon}}\approx 2.73\times10^{-3}\ \mathrm{m\,s^{-2}} \]

We can independently estimate the Moon’s centripetal acceleration using its orbital radius and period:

\[ a_{c}=\frac{4\pi^{2}r}{T^{2}} \]

Using rounded modern values:

• \(r=3.84\times10^{8}\ \mathrm{m}\), the Moon’s orbital radius.
• \(T=2.36\times10^{6}\ \mathrm{s}\), its orbital period.

Substitution gives:

\[ a_{c}= \frac{4\pi^{2}(3.84\times10^{8})} {(2.36\times10^{6})^{2}} \]

\[ a_{c}\approx2.73\times10^{-3}\ \mathrm{m\,s^{-2}} \]

The two estimates agree remarkably well.

This is a modern numerical reconstruction of the reasoning, not a reproduction of Newton’s original measurements. The historical achievement was recognising that one inverse-square gravitational model could explain both terrestrial falling motion and the Moon’s orbital acceleration.

It also reveals why a constant-\(g\) approximation cannot describe large changes in altitude. Gravity is not equally strong everywhere around Earth.

When does the constant-\(g\) approximation become unreliable?

The difference between the two models depends on the altitude relative to Earth’s radius.

Let \(R\) represent Earth’s radius and \(h\) the altitude above its surface. The distance from Earth’s centre is then:

\[ r=R+h \]

The gravitational acceleration at altitude \(h\) is:

\[ g_{h}=g_{0}\left(\frac{R}{R+h}\right)^{2} \]

Here, \(g_{0}\) is gravitational acceleration at Earth’s surface.

The following table uses illustrative rounded Earth values, with \(R=6370\ \mathrm{km}\).

The percentage overestimate is calculated relative to the exact increase in gravitational potential energy.

The pattern is clear. As altitude increases, the assumption of constant gravitational acceleration becomes less accurate.

For small changes in altitude, \(mgh\) remains useful. For large changes, we need an equation that accounts for the changing gravitational field throughout the journey.

03The exact gravitational potential energy equation

To understand the replacement for \(mgh\), we need to establish where gravitational potential energy is measured from.

For this explanation, gravitational potential energy is defined as zero at infinity.

That means:

\[ U=0 \qquad \text{at } r=\infty \]

At any finite distance from Earth’s centre, gravitational potential energy is negative:

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

The negative sign is important. It tells us that an object at a finite distance from Earth is in a gravitationally bound state. Energy must be supplied to move it to infinity and leave it with zero final speed.

Here:

• \(U\) is gravitational potential energy, in joules.
• \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\ \mathrm{N\,m^{2}\,kg^{-2}}\).
• \(M\) is Earth’s mass, in kilograms.
• \(m\) is the object’s mass, in kilograms.
• \(r\) is the distance from Earth’s centre, in metres.

Why is gravitational potential energy negative?

Think about an object initially at rest very far from Earth.

At infinity, its gravitational potential energy is zero. As it approaches Earth, gravity attracts it and can transfer energy into kinetic energy. Its gravitational potential energy decreases below zero.

The object has entered a gravitationally bound state.

This does not mean its kinetic energy is negative. Kinetic energy is:

\[ K=\frac12mv^{2} \]

It is always zero or positive.

The negative quantity is gravitational potential energy, not kinetic energy.

The distinction becomes especially important when discussing binding energy and escape. Binding energy is the positive amount of energy needed to take a gravitationally bound object to infinity with zero final speed. It is related to the object’s total mechanical energy, not automatically equal to the magnitude of its gravitational potential energy.

We’ll return to this distinction when considering satellites.

Deriving the exact change in gravitational potential energy

Suppose an object moves from Earth’s surface to an altitude \(h\).

Its initial distance from Earth’s centre is:

\[ r_{i}=R \]

Its final distance from Earth’s centre is:

\[ r_{f}=R+h \]

The change in gravitational potential energy is final energy minus initial energy:

\[ \Delta U=U_{f}-U_{i} \]

Substitute the gravitational potential energy equation:

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

Therefore:

\[ \boxed{ \Delta U=GMm\left( \frac{1}{R}-\frac{1}{R+h} \right) } \]

This is the exact increase in gravitational potential energy when an object moves from Earth’s surface to altitude \(h\), under the assumptions of a spherical Earth and Newtonian gravity.

Notice that the result is positive. Moving away from Earth increases gravitational potential energy, making it less negative.

The gravitational force does negative work during this outward movement. If an external force lifts the object slowly without changing its kinetic energy, that external force supplies the positive energy increase.

How does this connect back to \(mgh\)?

We can rewrite the exact equation using the surface gravitational acceleration:

\[ g_{0}=\frac{GM}{R^{2}} \]

Substituting \(GM=g_{0}R^{2}\) gives:

\[ \Delta U= \frac{g_{0}R^{2}mh}{R(R+h)} \]

Simplifying:

\[ \boxed{ \Delta U=mg_{0}h\left(\frac{R}{R+h}\right) } \]

Compare this with the familiar approximation:

\[ \Delta U_{\text{approx}}=mg_{0}h \]

The exact expression contains an additional factor:

\[ \frac{R}{R+h} \]

For a small altitude, \(h\) is much smaller than \(R\), so this factor is close to 1.

For a large altitude, the factor becomes significantly smaller than 1. The simple \(mgh\) equation therefore overestimates the energy required.

An especially useful result follows from comparing the two expressions:

\[ \frac{\Delta U_{\text{approx}}} {\Delta U_{\text{exact}}} =\frac{R+h}{R} \]

So the fractional overestimate, relative to the exact answer, is:

\[ \boxed{ \text{Fractional overestimate}=\frac{h}{R} } \]

For example, an altitude equal to 10% of Earth’s radius produces a 10% overestimate relative to the exact energy increase.

This gives you a quick way to judge whether the approximation is appropriate before reaching for your calculator.

04Worked examples

The exact equation is more useful once you can recognise when to use it. The following examples show two different situations: lifting an object through a large altitude change and calculating the energy of an object that is already gravitationally bound.

05Choosing the correct equation

There are several gravitational energy equations in HSC Physics. They answer different questions.

A reliable decision process is:

1. Identify whether the question asks for gravitational potential energy, a change in potential energy, work, or total mechanical energy.
2. Check whether the altitude change is small compared with the planet’s radius.
3. If using an inverse-square gravitational equation, convert altitude into distance from the centre using \(r=R+h\).
4. Establish the sign of the energy change.
5. If the object is moving, account for its kinetic energy before calculating binding energy.

One particularly important distinction: the exact change in gravitational potential energy is not automatically the change in total mechanical energy. If the object’s speed changes, its kinetic energy changes as well.

For a more detailed treatment of gravitational work between two orbital radii, see the separate calculation guide. The relationship between kinetic and potential energy in circular orbits is explained in How to Find Kinetic Energy from Potential Energy in a Circular Orbit.

06Questions and solutions

Question 1

A 250 kg equipment package is lifted from Earth’s surface to an altitude of 300 km.

Calculate the exact increase in gravitational potential energy.

Then calculate the percentage by which \(mgh\), using \(g=9.81\ \mathrm{m\,s^{-2}}\), overestimates the exact result.

Use \(R=6.37\times10^{6}\ \mathrm{m}\) and \(GM=3.98\times10^{14}\ \mathrm{m^{3}\,s^{-2}}\).

Solution 1

The exact increase in gravitational potential energy is approximately \(7.03\times10^{8}\ \mathrm{J}\), and the \(mgh\) approximation overestimates this by approximately 4.7%.

Step 1: Identify the initial and final radii.

The initial radius is:

\[ r_{i}=6.37\times10^{6}\ \mathrm{m} \]

The final radius is:

\[ r_{f}=6.37\times10^{6}+3.00\times10^{5} \]

\[ r_{f}=6.67\times10^{6}\ \mathrm{m} \]

Step 2: Calculate the exact change in gravitational potential energy.

\[ \Delta U=GMm\left( \frac{1}{r_{i}}-\frac{1}{r_{f}} \right) \]

Substitute:

\[ \Delta U=(3.98\times10^{14})(250) \left( \frac{1}{6.37\times10^{6}} -\frac{1}{6.67\times10^{6}} \right) \]

\[ \boxed{\Delta U=7.03\times10^{8}\ \mathrm{J}} \]

Step 3: Calculate the approximate result using \(mgh\).

\[ \Delta U_{\text{approx}}=mgh \]

\[ =250(9.81)(3.00\times10^{5}) \]

\[ =7.36\times10^{8}\ \mathrm{J} \]

Step 4: Calculate the percentage overestimate.

\[ \text{Overestimate}= \frac{\Delta U_{\text{approx}}-\Delta U} {\Delta U}\times100\% \]

\[ =\frac{7.36-7.03}{7.03}\times100\% \]

\[ \boxed{\text{Overestimate}=4.7\%} \]

The approximation is reasonably close in this situation, but the exact equation is still more accurate.

The error comes from treating gravitational acceleration as constant when it actually decreases with altitude.

Question 2

A 750 kg satellite is transferred from an altitude of 400 km to an altitude of 1,600 km above Earth’s surface.

Calculate the increase in its gravitational potential energy.

A student uses 400 km and 1,600 km directly as the initial and final radii in the gravitational potential energy equation.

Explain why this is incorrect.

Use:

• \(R=6.37\times10^{6}\ \mathrm{m}\)
• \(GM=3.98\times10^{14}\ \mathrm{m^{3}\,s^{-2}}\)

Assume the satellite’s final and initial positions are measured relative to Earth’s centre, and ignore other gravitational influences.

Solution 2

The satellite’s gravitational potential energy increases by approximately \(6.64\times10^{9}\ \mathrm{J}\). The student’s method is incorrect because orbital radius is measured from Earth’s centre, not from its surface.

Step 1: Convert both altitudes into metres.

\[ h_{i}=4.00\times10^{5}\ \mathrm{m} \]

\[ h_{f}=1.60\times10^{6}\ \mathrm{m} \]

Step 2: Calculate the actual initial and final radii.

\[ r_{i}=R+h_{i} \]

\[ r_{i}=6.77\times10^{6}\ \mathrm{m} \]

Similarly:

\[ r_{f}=R+h_{f} \]

\[ r_{f}=7.97\times10^{6}\ \mathrm{m} \]

Step 3: Apply the exact potential energy change equation.

\[ \Delta U=GMm\left( \frac{1}{r_{i}}-\frac{1}{r_{f}} \right) \]

Substitute:

\[ \Delta U=(3.98\times10^{14})(750) \left( \frac{1}{6.77\times10^{6}} -\frac{1}{7.97\times10^{6}} \right) \]

\[ \boxed{\Delta U=6.64\times10^{9}\ \mathrm{J}} \]

Step 4: Explain the student’s mistake.

The student has confused altitude with orbital radius.

Altitude is measured from Earth’s surface, while orbital radius is measured from Earth’s centre.

The correct relationship is:

\[ \boxed{r=R+h} \]

Using the altitudes directly would produce an incorrect result of approximately \(5.60\times10^{11}\ \mathrm{J}\), vastly larger than the correct answer.

The key lesson is to establish the distance from the centre before substituting into any inverse-square gravitational equation.

Question 3

A student calculates the gravitational potential energy increase for an object lifted from Earth’s surface using \(mgh\).

The calculated value is 12% greater than the exact increase in gravitational potential energy.

Determine:

(a) The altitude to which the object was lifted.

(b) The gravitational acceleration at that altitude, expressed as a fraction of its surface value.

Use Earth’s radius as \(6,370\ \mathrm{km}\).

Solution 3

The object was lifted approximately 764 km above Earth’s surface, where gravitational acceleration is approximately \(0.797g_{0}\), or 79.7% of its surface value.

Step 1: Relate the percentage error to altitude.

We established that:

\[ \frac{\Delta U_{\text{approx}}} {\Delta U_{\text{exact}}} =1+\frac{h}{R} \]

A 12% overestimate means:

\[ \frac{\Delta U_{\text{approx}}} {\Delta U_{\text{exact}}}=1.12 \]

Therefore:

\[ 1.12=1+\frac{h}{R} \]

Rearrange:

\[ \frac{h}{R}=0.12 \]

Step 2: Calculate the altitude.

\[ h=0.12R \]

\[ h=0.12(6370) \]

\[ \boxed{h=764.4\ \mathrm{km}} \]

Step 3: Calculate gravitational acceleration at this altitude.

Use:

\[ g_{h}=g_{0}\left(\frac{R}{R+h}\right)^{2} \]

Substitute the altitude:

\[ g_{h}=g_{0}\left( \frac{6370}{6370+764.4} \right)^{2} \]

\[ g_{h}=g_{0}\left(\frac{1}{1.12}\right)^{2} \]

\[ \boxed{g_{h}=0.797g_{0}} \]

Step 4: Interpret the result.

Gravitational acceleration has decreased by approximately 20.3%, even though the \(mgh\) approximation overestimates the exact energy increase by 12%.

These percentages are not interchangeable.

The energy error depends linearly on \(h/R\), while gravitational acceleration follows an inverse-square relationship.

This is why you should not assume that a 12% error in gravitational potential energy means gravitational acceleration has decreased by 12%.

Question 4

A 500 kg satellite is travelling in a circular orbit around Earth. Its gravitational potential energy, measured relative to zero at infinity, is:

\[ U=-1.20\times10^{10}\ \mathrm{J} \]

An engine then supplies \(7.0\times10^{9}\ \mathrm{J}\) of additional mechanical energy.

Determine:

(a) The satellite’s initial kinetic energy.

(b) Its initial total mechanical energy and binding energy.

(c) Whether the satellite can escape Earth after the engine burn.

(d) Its speed at infinity after escaping.

Assume the engine’s work is transferred entirely into the satellite’s mechanical energy, with no losses or other external influences.

Solution 4

The satellite initially has a kinetic energy of \(6.0\times10^{9}\ \mathrm{J}\), a binding energy of \(6.0\times10^{9}\ \mathrm{J}\), and escapes after the engine supplies \(7.0\times10^{9}\ \mathrm{J}\). Its speed at infinity is \(2,000\ \mathrm{m\,s^{-1}}\).

Step 1: Calculate the initial kinetic energy.

For a circular orbit:

\[ K=-\frac{U}{2} \]

Substitute:

\[ K=-\frac{-1.20\times10^{10}}{2} \]

\[ \boxed{K=6.0\times10^{9}\ \mathrm{J}} \]

Step 2: Calculate the initial total mechanical energy.

\[ E=K+U \]

\[ E=6.0\times10^{9}-1.20\times10^{10} \]

\[ \boxed{E=-6.0\times10^{9}\ \mathrm{J}} \]

The negative total energy indicates that the satellite is gravitationally bound.

Its binding energy is:

\[ B=-E \]

\[ \boxed{B=6.0\times10^{9}\ \mathrm{J}} \]

Step 3: Determine whether the engine supplies enough energy to escape.

The engine adds \(7.0\times10^{9}\ \mathrm{J}\).

Therefore:

\[ E_{f}=E_{i}+\Delta E_{\text{engine}} \]

\[ E_{f}=-6.0\times10^{9}+7.0\times10^{9} \]

\[ \boxed{E_{f}=+1.0\times10^{9}\ \mathrm{J}} \]

The final total mechanical energy is positive.

The satellite is now unbound and can escape to infinity.

Notice that the engine did not need to supply the magnitude of the original gravitational potential energy. It only needed to overcome the initial binding energy.

Step 4: Calculate the speed at infinity.

At infinity:

\[ U_{\infty}=0 \]

Therefore, the remaining total mechanical energy becomes kinetic energy:

\[ E_{f}=K_{\infty} \]

\[ 1.0\times10^{9}=\frac12mv_{\infty}^{2} \]

Rearrange:

\[ v_{\infty}=\sqrt{\frac{2E_{f}}{m}} \]

Substitute:

\[ v_{\infty}= \sqrt{\frac{2(1.0\times10^{9})}{500}} \]

\[ \boxed{v_{\infty}=2.00\times10^{3}\ \mathrm{m\,s^{-1}}} \]

The satellite escapes with a residual speed of 2,000 m/s.

This is another important distinction: escaping does not necessarily mean reaching infinity with zero speed. That only happens when the final total mechanical energy is exactly zero.

Question 5

Two identical probes, each of mass \(m\), are located at the same distance \(r\) from the centre of an isolated planet.

Their initial conditions are different:

• Probe A is momentarily stationary.
• Probe B is travelling directly away from the planet at \(0.80v_{\text{esc}}\).

Here, \(v_{\text{esc}}\) is the escape speed at radius \(r\):

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

Both probes receive an identical short outward impulse. Their speeds each increase by \(0.30v_{\text{esc}}\).

Assume:

• Both probes experience the same gravitational potential energy at the start of the impulse.
• The impulse changes their speeds without appreciably changing their positions.
• No other forces act, and the planet’s mass is much greater than either probe’s mass.

Determine:

(a) The final total mechanical energy of each probe, expressed as a multiple of \(GMm/r\).

(b) The work done by the engine on each probe, expressed as a multiple of \(GMm/r\).

(c) Which probe escapes and which remains gravitationally bound.

(d) The speed of the escaping probe at infinity, expressed as a multiple of its original local escape speed.

Explain why giving both probes the same increase in speed does not mean giving them the same amount of energy.

Solution 5

Probe A remains gravitationally bound, with negative total mechanical energy. Probe B escapes, with positive total mechanical energy. Despite receiving the same increase in speed, probe B receives more than six times as much energy from the impulse.

Step 1: Establish the gravitational potential energy shared by both probes.

Both probes have the same mass and are at the same radius.

Therefore:

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

This potential energy does not change significantly during the short impulse.

Step 2: Calculate the final speed of each probe.

Probe A starts from rest:

\[ v_{A}=0+0.30v_{\text{esc}} \]

\[ v_{A}=0.30v_{\text{esc}} \]

Probe B already has an outward speed:

\[ v_{B}=0.80v_{\text{esc}}+0.30v_{\text{esc}} \]

\[ v_{B}=1.10v_{\text{esc}} \]

The probes have different final speeds, even though their speed increases were identical.

Step 3: Calculate the final total mechanical energy of Probe A.

Recall:

\[ E=K+U \]

For Probe A:

\[ E_{A}=\frac12m(0.30v_{\text{esc}})^{2} -\frac{GMm}{r} \]

Expand:

\[ E_{A}=0.045mv_{\text{esc}}^{2} -\frac{GMm}{r} \]

Using:

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

We obtain:

\[ E_{A}=0.045m\left(\frac{2GM}{r}\right) -\frac{GMm}{r} \]

Therefore:

\[ \boxed{E_{A}=-0.91\frac{GMm}{r}} \]

Its total mechanical energy is negative, so Probe A remains gravitationally bound.

Step 4: Calculate the final total mechanical energy of Probe B.

Its final speed is \(1.10v_{\text{esc}}\).

\[ E_{B}=\frac12m(1.10v_{\text{esc}})^{2} -\frac{GMm}{r} \]

Expand:

\[ E_{B}=0.605mv_{\text{esc}}^{2} -\frac{GMm}{r} \]

Substitute the escape speed relationship:

\[ E_{B}=0.605m\left(\frac{2GM}{r}\right) -\frac{GMm}{r} \]

Therefore:

\[ \boxed{E_{B}=+0.21\frac{GMm}{r}} \]

Its total mechanical energy is positive, so Probe B escapes.

Step 5: Calculate the work done on each probe.

During the short impulse, the change in gravitational potential energy is negligible.

Therefore, the work supplied by the engine equals the change in kinetic energy:

\[ W=\Delta K \]

For Probe A, the initial speed is zero:

\[ W_{A}=\frac12m(0.30v_{\text{esc}})^{2} \]

\[ W_{A}=0.045mv_{\text{esc}}^{2} \]

Substituting the escape speed equation:

\[ \boxed{W_{A}=0.09\frac{GMm}{r}} \]

For Probe B, we must subtract its initial kinetic energy from its final kinetic energy:

\[ W_{B}=\frac12m\left[ (1.10v_{\text{esc}})^{2} -(0.80v_{\text{esc}})^{2} \right] \]

Simplifying:

\[ W_{B}=\frac12m(1.21-0.64)v_{\text{esc}}^{2} \]

\[ W_{B}=0.285mv_{\text{esc}}^{2} \]

Therefore:

\[ \boxed{W_{B}=0.57\frac{GMm}{r}} \]

Probe B receives more energy because it was already moving when the impulse was applied.

Step 6: Calculate Probe B’s speed at infinity.

At infinity, gravitational potential energy is zero.

Therefore:

\[ E_{B}=\frac12mv_{\infty}^{2} \]

Substitute the final energy:

\[ 0.21\frac{GMm}{r} =\frac12mv_{\infty}^{2} \]

Rearrange:

\[ v_{\infty}^{2}=0.42\frac{GM}{r} \]

Recall that:

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

Therefore:

\[ v_{\infty}^{2}=0.21v_{\text{esc}}^{2} \]

Taking the square root:

\[ \boxed{v_{\infty}=\sqrt{0.21}\,v_{\text{esc}}} \]

\[ \boxed{v_{\infty}\approx0.458v_{\text{esc}}} \]

Probe B escapes with approximately 45.8% of its original local escape speed remaining at infinity.

Why is this result surprising?

You might initially expect identical impulses to transfer identical amounts of energy.

That is incorrect because impulse and work describe different things.

The impulse determines the change in momentum. Work determines the change in kinetic energy.

For a mass receiving a speed increase \(\Delta v\):

\[ \Delta K= \frac12m\left[(v+\Delta v)^{2}-v^{2}\right] \]

Expanding:

\[ \Delta K=mv\Delta v+\frac12m(\Delta v)^{2} \]

The first term depends on the initial speed.

Probe A starts from rest, so this term is zero. Probe B already has substantial outward velocity, so the same increase in speed produces a much larger increase in kinetic energy.

That is why Probe B escapes while Probe A remains bound.

The result also reinforces the importance of total mechanical energy. Knowing the change in speed alone is not enough to determine whether an object can escape a gravitational field.

07Where to go next

The key step is recognising that \(mgh\) is a local approximation, not a universal gravitational potential energy equation.

For large changes in altitude, use:

\[ \boxed{ \Delta U=GMm\left( \frac{1}{r_{i}}-\frac{1}{r_{f}} \right) } \]

Keep gravitational potential energy zero at infinity, remember that kinetic energy is always non-negative, and use total mechanical energy to determine whether an object is gravitationally bound.

The natural next step is understanding how this energy relationship determines escape speed, including why an object needs a particular initial speed to reach infinity with zero remaining kinetic energy.