How to Calculate Gravitational Work Between Two Orbital Radii

Suppose a 750 kg satellite moves from an orbital radius of \(7.00\times10^6\) m to \(1.40\times10^7\) m around Earth. Gravity pulls the satellite towards Earth throughout the journey, but the satellite moves outwards. Does gravity do positive or negative work?

The answer is negative work. However, calculating its magnitude requires more than multiplying gravitational force by distance. The gravitational force decreases as the satellite moves further from Earth, so we must account for that changing force.

The formula you need is:

\[ \boxed{W_g=GMm\left(\frac{1}{r_f}-\frac{1}{r_i}\right)} \]

This calculates the work done by gravity when an object moves between two distances from the centre of a massive body.

Let’s build up where the formula comes from, how to use it in HSC Physics calculations, and why gravitational work is not necessarily the same as the change in kinetic energy or the energy needed to escape an orbit.

01Understanding gravitational work between two radii

First, identify the quantities involved.

For Earth, you will commonly use:

\[ G=6.67\times10^{-11} \]

\[ M_E=5.97\times10^{24}\ \mathrm{kg} \]

It is useful to combine these two constants:

\[ \mu=GM \]

For Earth:

\[ \mu_E=3.98\times10^{14}\ \mathrm{m^3\,s^{-2}} \]

This is simply a mathematical shortcut. It does not introduce a different gravitational law.

Be careful: orbital radius is not altitude

This is one of the easiest places to lose marks.

Orbital radius is measured from the centre of the planet, whereas altitude is measured from its surface.

If a question gives the satellite’s altitude \(h\), you must calculate:

\[ r=R+h \]

where \(R\) is the radius of the planet.

For example, a satellite 400 km above Earth’s surface has an orbital radius of approximately:

\[ r=6.37\times10^6+4.00\times10^5 \]

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

Putting 400 km directly into the gravitational work equation would give a completely incorrect result.

02Why does the gravitational work equation look like this?

You already know that work is calculated using force and displacement.

For a constant force acting in the direction of displacement:

\[ W=Fs \]

But gravitational force is not constant over large orbital distances.

Newton’s law of universal gravitation gives:

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

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

So, when a satellite moves between two orbital radii, we cannot simply multiply the initial gravitational force by the total displacement.

Instead, we need to account for the changing force throughout the journey.

Deriving the gravitational work equation

Take the outward radial direction as positive.

Gravity acts towards the central body, so its radial force component is:

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

For a very small radial displacement \(dr\), the work done by gravity is:

\[ dW_g=F_r\,dr \]

Substituting the gravitational force:

\[ dW_g=-\frac{GMm}{r^2}\,dr \]

To calculate the total work, integrate from the initial radius to the final radius:

\[ W_g=-GMm\int_{r_i}^{r_f}r^{-2}\,dr \]

Integrating:

\[ W_g=-GMm\left[-\frac{1}{r}\right]_{r_i}^{r_f} \]

Therefore:

\[ \boxed{W_g=GMm\left(\frac{1}{r_f}-\frac{1}{r_i}\right)} \]

This is the equation to use when calculating gravitational work between two orbital radii.

An important detail is that gravitational force is conservative. Its work depends only on the initial and final positions, not the path taken between them.

Even if a satellite follows a curved orbital trajectory, the tangential component of its displacement does no work against the radial gravitational force.

What determines the sign of the work?

Before substituting numbers, compare the initial and final radii.

Why?

• When a satellite moves outwards, its displacement is opposite to gravitational force. Gravity does negative work.
• When a satellite moves inwards, its displacement is in the same direction as gravitational force. Gravity does positive work.

This sign check is worth doing before every calculation. It catches many substitution errors.

03Gravitational work and gravitational potential energy

There is another quantity that often appears alongside gravitational work: gravitational potential energy.

With gravitational potential energy defined as zero at infinity:

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

The negative sign tells us that an object at a finite distance from the central body has less gravitational potential energy than it would have at infinity.

The change in gravitational potential energy is:

\[ \Delta U=U_f-U_i \]

Substituting the potential energy equation:

\[ \Delta U=-\frac{GMm}{r_f} +\frac{GMm}{r_i} \]

Rearranging:

\[ \boxed{\Delta U=GMm\left(\frac{1}{r_i}-\frac{1}{r_f}\right)} \]

Compare this with the gravitational work equation:

\[ \boxed{W_g=-\Delta U} \]

This relationship is extremely important.

The work done by gravity is the negative of the change in gravitational potential energy.

For example, when a satellite moves outwards:

• Gravity does negative work.
• Gravitational potential energy increases.
• When gravity is the only force doing work, kinetic energy decreases.

The signs are not contradictory. They describe different energy changes.

A common misconception: negative potential energy means negative kinetic energy

No. These are different quantities.

Gravitational potential energy is negative when we choose zero potential energy at infinity. Kinetic energy is always zero or positive.

For a satellite in a circular orbit, three energy quantities are particularly important.

Kinetic energy:

\[ K=\frac12mv^2 \]

Gravitational potential energy:

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

Total mechanical orbital energy:

\[ E=K+U \]

For a circular orbit, gravitational force provides the centripetal force:

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

Rearranging:

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

Substitute this into the kinetic energy equation:

\[ K=\frac{GMm}{2r} \]

Therefore, the total mechanical energy is:

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

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

The negative total energy tells us that the satellite is gravitationally bound. It does not have enough total mechanical energy to escape to infinity.

What is binding energy?

Binding energy is the energy required to take an object from its bound orbital state to infinity with zero final speed.

It is the magnitude of the negative total mechanical energy:

\[ B=-E \]

For a circular orbit:

\[ \boxed{B=\frac{GMm}{2r}} \]

Notice something interesting:

\[ \boxed{B=K} \]

For a circular orbit, binding energy and kinetic energy have the same numerical value.

However, they are not the same physical quantity.

• Kinetic energy is the energy associated with motion.
• Binding energy is the energy required to remove the object from its gravitationally bound state.

This distinction becomes especially important in escape velocity and orbital transfer questions.

04A reliable method for HSC calculations

When you see a question asking for gravitational work between two orbital radii, follow this process.

Step 1: Identify the central body and orbiting object.

Determine \(M\) and \(m\).

Step 2: Convert both radii into metres.

If altitudes are provided, add the planet’s radius.

Step 3: Identify the requested quantity.

Is the question asking for:

• Work done by gravity?
• Change in gravitational potential energy?
• Change in kinetic energy?
• Work required to escape?
• Change in total orbital energy?

These are related, but they are not interchangeable.

Step 4: Select the correct equation.

For work done by gravity:

\[ W_g=GMm\left(\frac{1}{r_f}-\frac{1}{r_i}\right) \]

For the change in gravitational potential energy:

\[ \Delta U=-W_g \]

For the change in kinetic energy:

\[ \Delta K=W_{\text{net}} \]

Remember that the net work includes the work done by all forces, not just gravity.

Step 5: Check the sign and units.

Your answer should be in joules.

An outward movement should produce negative gravitational work. An inward movement should produce positive gravitational work.

05Worked examples

06Questions and solutions

Use:

\[ \mu_E=3.98\times10^{14}\ \mathrm{m^3\,s^{-2}} \]

for all questions involving Earth.

Question 1

A 500 kg satellite moves from an orbital radius of \(8.00\times10^6\) m to \(1.00\times10^7\) m around Earth.

Calculate:

a) The work done by gravity.

b) The change in gravitational potential energy.

Solution 1

The work done by gravity is \(-4.98\times10^9\) J, and gravitational potential energy increases by \(+4.98\times10^9\) J.

Part a: Work done by gravity

Use:

\[ W_g=\mu_E m \left(\frac{1}{r_f}-\frac{1}{r_i}\right) \]

Substitute:

\[ \begin{aligned} W_g&=(3.98\times10^{14})(500)\\ &\quad\times\left( \frac{1}{1.00\times10^7} -\frac{1}{8.00\times10^6} \right)\\ &=(1.99\times10^{17})(-2.50\times10^{-8})\\ &=-4.98\times10^9\ \mathrm{J} \end{aligned} \]

Part b: Change in gravitational potential energy

Since:

\[ \Delta U=-W_g \]

\[ \boxed{\Delta U=+4.98\times10^9\ \mathrm{J}} \]

The satellite moves outwards, so gravity does negative work while gravitational potential energy increases.

Question 2

An 850 kg probe moves inwards towards Earth, from an initial orbital radius of \(2.40\times10^7\) m to a final radius of \(8.00\times10^6\) m.

Calculate:

a) The work done by gravity.

b) The change in gravitational potential energy.

c) The change in kinetic energy, assuming gravity is the only force doing work.

d) Explain why the initial and final orbital radii alone are insufficient to determine the probe’s final speed.

Solution 2

Gravity does \(+2.82\times10^{10}\) J of work. Potential energy decreases by the same magnitude, and kinetic energy increases by \(2.82\times10^{10}\) J if gravity is the only force doing work.

Part a: Work done by gravity

\[ W_g=\mu_E m \left(\frac{1}{r_f}-\frac{1}{r_i}\right) \]

Substitute:

\[ \begin{aligned} W_g&=(3.98\times10^{14})(850)\\ &\quad\times\left( \frac{1}{8.00\times10^6} -\frac{1}{2.40\times10^7} \right)\\ &=(3.383\times10^{17})(8.33\times10^{-8})\\ &=+2.82\times10^{10}\ \mathrm{J} \end{aligned} \]

The positive sign is expected because gravity and displacement are both directed inwards.

Part b: Change in gravitational potential energy

\[ \Delta U=-W_g \]

\[ \boxed{\Delta U=-2.82\times10^{10}\ \mathrm{J}} \]

Part c: Change in kinetic energy

The work-energy theorem states:

\[ W_{\text{net}}=\Delta K \]

Since gravity is the only force doing work:

\[ \boxed{\Delta K=+2.82\times10^{10}\ \mathrm{J}} \]

The probe gains kinetic energy as it moves inwards.

Part d: Can we determine the final speed?

No. We know the change in kinetic energy, but we do not know the initial kinetic energy.

Using:

\[ \Delta K=\frac12m(v_f^2-v_i^2) \]

we can determine:

\[ v_f^2-v_i^2=\frac{2W_g}{m} \]

However, without \(v_i\), there is no unique value for \(v_f\).

This is an important distinction: knowing the work done tells us how much kinetic energy changes, not necessarily the final speed.

Question 3

A 600 kg satellite follows an unpowered elliptical orbit around Earth.

Its perigee is at a radius of \(7.00\times10^6\) m, and its apogee is at a radius of \(2.10\times10^7\) m.

Calculate:

a) The work done by gravity as the satellite travels from perigee to apogee.

b) The change in gravitational potential energy.

c) The change in kinetic energy.

Explain why the satellite’s speed decreases during this part of its orbit, even though it continues moving along its orbital path.

Solution 3

Gravity does \(-2.27\times10^{10}\) J of work. Gravitational potential energy increases by \(+2.27\times10^{10}\) J, and kinetic energy decreases by \(2.27\times10^{10}\) J.

Part a: Work done by gravity

The satellite moves between two radial distances, so the same gravitational work equation applies.

\[ W_g=\mu_E m \left(\frac{1}{r_f}-\frac{1}{r_i}\right) \]

Substitute:

\[ \begin{aligned} W_g&=(3.98\times10^{14})(600)\\ &\quad\times\left( \frac{1}{2.10\times10^7} -\frac{1}{7.00\times10^6} \right)\\ &=(2.388\times10^{17})(-9.52\times10^{-8})\\ &=-2.27\times10^{10}\ \mathrm{J} \end{aligned} \]

Part b: Change in gravitational potential energy

\[ \Delta U=-W_g \]

\[ \boxed{\Delta U=+2.27\times10^{10}\ \mathrm{J}} \]

Part c: Change in kinetic energy

The satellite is on an unpowered orbit, so gravity is the only force doing work.

Therefore:

\[ \Delta K=W_g \]

\[ \boxed{\Delta K=-2.27\times10^{10}\ \mathrm{J}} \]

The satellite loses kinetic energy as it moves from perigee to apogee.

Why does this happen?

Gravity acts towards Earth, while the satellite moves towards a greater orbital radius.

Although the satellite also moves tangentially along its elliptical path, the gravitational force acts radially. The tangential component of displacement does no work.

The outward radial movement means gravity does negative work, reducing the satellite’s kinetic energy and increasing its gravitational potential energy.

The satellite is slowing down as it approaches apogee. It will speed up again as it falls back towards perigee.

Question 4

A 900 kg satellite is transferred from an initial circular orbit of radius \(8.00\times10^6\) m to a final circular orbit of radius \(1.60\times10^7\) m around Earth.

Assume there is no atmospheric drag or other dissipative force.

Calculate:

a) The work done by gravity during the transfer.

b) The change in gravitational potential energy.

c) The initial and final kinetic energies.

d) The net work done by the propulsion system.

e) Explain why the work done by gravity is not equal to the change in total orbital energy.

Solution 4

Gravity does \(-2.24\times10^{10}\) J of work, while the propulsion system does a net \(+1.12\times10^{10}\) J of work. The satellite’s total mechanical orbital energy increases by \(1.12\times10^{10}\) J.

Part a: Work done by gravity

Use:

\[ W_g=\mu_E m \left(\frac{1}{r_f}-\frac{1}{r_i}\right) \]

Substitute:

\[ \begin{aligned} W_g&=(3.98\times10^{14})(900)\\ &\quad\times\left( \frac{1}{1.60\times10^7} -\frac{1}{8.00\times10^6} \right)\\ &=(3.582\times10^{17})(-6.25\times10^{-8})\\ &=-2.24\times10^{10}\ \mathrm{J} \end{aligned} \]

Part b: Change in gravitational potential energy

\[ \Delta U=-W_g \]

\[ \boxed{\Delta U=+2.24\times10^{10}\ \mathrm{J}} \]

Part c: Initial and final kinetic energies

Because both orbits are circular, use:

\[ K=\frac{GMm}{2r} \]

For the initial orbit:

\[ \begin{aligned} K_i&=\frac{(3.98\times10^{14})(900)} {2(8.00\times10^6)}\\ &=2.24\times10^{10}\ \mathrm{J} \end{aligned} \]

For the final orbit:

\[ \begin{aligned} K_f&=\frac{(3.98\times10^{14})(900)} {2(1.60\times10^7)}\\ &=1.12\times10^{10}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \Delta K=K_f-K_i \]

\[ \boxed{\Delta K=-1.12\times10^{10}\ \mathrm{J}} \]

The satellite has less kinetic energy in the higher circular orbit because its required circular orbital speed is lower.

Part d: Net work done by propulsion

The work-energy theorem gives:

\[ W_{\text{net}}=\Delta K \]

The net work includes gravity and propulsion:

\[ W_g+W_{\text{propulsion}}=\Delta K \]

Rearranging:

\[ W_{\text{propulsion}}=\Delta K-W_g \]

Substitute:

\[ \begin{aligned} W_{\text{propulsion}} &=-1.12\times10^{10}\\ &\quad-(-2.24\times10^{10})\\ &=+1.12\times10^{10}\ \mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{W_{\text{propulsion}}=+1.12\times10^{10}\ \mathrm{J}} \]

Part e: Why isn’t gravitational work equal to the change in total orbital energy?

Because gravitational work changes the satellite’s kinetic energy, while the gravitational potential energy changes in the opposite direction.

The change in total mechanical energy is:

\[ \Delta E=\Delta K+\Delta U \]

Substitute:

\[ \begin{aligned} \Delta E&=-1.12\times10^{10}\\ &\quad+2.24\times10^{10}\\ &=+1.12\times10^{10}\ \mathrm{J} \end{aligned} \]

This equals the net work done by propulsion.

Gravity is a conservative force, so its work is already accounted for through the change in gravitational potential energy. The propulsion system supplies the increase in total mechanical energy.

The trap is to assume that because gravity does negative work, the satellite’s total orbital energy must also decrease. That would ignore the additional work supplied by the propulsion system.

Question 5

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

Its initial orbital speed is:

\[ v_0=\sqrt{\frac{GM}{r}} \]

A propulsion system guides the satellite outwards to a distance of \(2r\), maintaining its speed at exactly \(v_0\) throughout the journey.

At the final position, the satellite’s velocity is tangential to the circular path at radius \(2r\). The propulsion system then switches off.

Assume Newtonian gravity and no other forces.

Determine:

a) The work done by gravity during the outward movement.

b) The net work done by the propulsion system.

c) Whether the satellite is in a circular orbit, a bound elliptical orbit, or an escape trajectory after the propulsion system switches off.

d) Explain the relationship between the propulsion work and the satellite’s initial binding energy.

Solution 5

The work done by gravity is \(-GMm/(2r)\), while the propulsion system supplies \(+GMm/(2r)\). When the propulsion switches off, the satellite is at the exact escape speed at radius \(2r\), so it follows a parabolic escape trajectory.

This question combines gravitational work, kinetic energy, circular orbital speed, and binding energy.

Part a: Work done by gravity

Apply the standard equation:

\[ W_g=GMm\left(\frac{1}{r_f}-\frac{1}{r_i}\right) \]

The initial and final radii are:

\[ r_i=r,\qquad r_f=2r \]

Therefore:

\[ \begin{aligned} W_g&=GMm\left(\frac{1}{2r}-\frac{1}{r}\right)\\ &=GMm\left(-\frac{1}{2r}\right) \end{aligned} \]

Hence:

\[ \boxed{W_g=-\frac{GMm}{2r}} \]

Gravity does negative work because the satellite moves outwards.

Part b: Work done by propulsion

The satellite maintains a constant speed throughout the journey.

Since kinetic energy depends on speed:

\[ K=\frac12mv^2 \]

its initial and final kinetic energies are equal.

Therefore:

\[ \Delta K=0 \]

Apply the work-energy theorem:

\[ W_{\text{net}}=\Delta K \]

\[ W_g+W_{\text{propulsion}}=0 \]

Rearranging:

\[ W_{\text{propulsion}}=-W_g \]

Therefore:

\[ \boxed{W_{\text{propulsion}}=+\frac{GMm}{2r}} \]

The propulsion system must supply positive work to compensate for the negative work done by gravity.

A tempting mistake is to assume that constant speed means no work is being done. It means the net work is zero, not that each individual force does zero work.

Part c: What happens when propulsion switches off?

This is the key part of the question.

The satellite’s initial circular orbital speed is:

\[ v_0=\sqrt{\frac{GM}{r}} \]

The propulsion system maintains this speed, so its speed at radius \(2r\) is still:

\[ v_f=v_0 \]

But is this the correct speed for a circular orbit at \(2r\)?

The circular orbital speed at that radius would be:

\[ v_c=\sqrt{\frac{GM}{2r}} \]

Rearranging:

\[ v_c=\frac{v_0}{\sqrt{2}} \]

The satellite is therefore travelling faster than the circular orbital speed at its final radius.

However, this alone does not establish whether it escapes. We need to compare its speed with the escape speed.

The escape speed at radius \(2r\) is:

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

Simplifying:

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

But this is exactly the initial circular orbital speed:

\[ \boxed{v_{\text{esc}}=v_0} \]

The satellite reaches the exact escape speed.

Therefore, when propulsion switches off, its total mechanical energy is zero, and it follows a parabolic escape trajectory.

It is neither in a circular orbit nor in a bound elliptical orbit.

Part d: Relationship with initial binding energy

Initially, the satellite is in a circular orbit.

Its initial binding energy is:

\[ B=\frac{GMm}{2r} \]

From part b, the propulsion system supplies:

\[ W_{\text{propulsion}}=\frac{GMm}{2r} \]

Therefore:

\[ \boxed{W_{\text{propulsion}}=B} \]

The propulsion system supplies exactly the energy needed to remove the satellite’s initial gravitational binding.

This is why the satellite reaches the escape threshold.

The subtle point is that the satellite never needed to increase its speed to gain enough energy to escape. Its speed remained constant, but its gravitational potential energy increased as it moved outwards.

That is the difference between tracking kinetic energy and tracking the total mechanical energy of an orbiting object.

07Where to go next

Once you can calculate gravitational work between two radii, the natural next step is understanding how orbital speed changes with radius.

For a circular orbit:

\[ v=\sqrt{\frac{GM}{r}} \]

This explains why satellites in higher circular orbits move more slowly, even though they have greater gravitational potential energy.

Work, potential energy, kinetic energy, and orbital velocity are different ways of describing the same gravitational system. Being able to move between them is particularly useful for longer HSC questions involving satellite transfers and escape velocity.

Continue with Orbital Velocity: Derivation and Calculations for HSC Physics.