How to Calculate the Energy Needed to Raise a Satellite’s Orbit

A satellite in a higher circular orbit moves more slowly than it did in its original orbit. If you judge its energy by its speed alone, you might predict that raising the orbit requires the satellite to lose energy.

That prediction is wrong. Raising a satellite into a higher circular orbit requires a positive increase in its total orbital mechanical energy, even though its kinetic energy decreases.

The calculation is based on the difference between the total energies of the initial and final circular orbits. Once you understand why, you can calculate the energy required for almost any change in circular orbital radius.

01The energy calculation: what are we actually finding?

Imagine a satellite moving around Earth. Its engine fires, and the satellite eventually settles into a higher circular orbit.

Predict what happens to each of these quantities:

• Kinetic energy
• Gravitational potential energy
• Total mechanical energy

The satellite moves more slowly in its higher circular orbit, so its kinetic energy decreases. However, it is also farther from Earth, so its gravitational potential energy increases.

The increase in gravitational potential energy is twice the decrease in kinetic energy. The difference is the positive increase in total mechanical energy.

Start with the energy of a circular orbit

We choose gravitational potential energy to be zero when the satellite and Earth are infinitely far apart.

This gives gravitational potential energy at a distance \(r\):

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

Here:

• \(U\) is gravitational potential energy, in joules (J).
• \(G\) 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 satellite’s mass, in kilograms.
• \(r\) is the distance between the centres of Earth and the satellite, in metres.

The negative sign is important. A satellite in orbit has negative gravitational potential energy because it is gravitationally bound to Earth.

Its kinetic energy, however, is always positive:

\[ K=\frac12 mv^2 \]

For a circular orbit, gravity provides the centripetal force required to maintain the satellite’s circular motion.

Equating gravitational force and centripetal force:

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

Rearranging:

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

Substituting this into the kinetic energy equation:

\[ K=\frac12m\left(\frac{GM}{r}\right) \]

Therefore:

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

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

\[ E=K+U \]

Substituting both expressions:

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

Therefore, for a circular orbit:

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

This is the equation you need to calculate the energy required to raise a satellite’s circular orbit.

Why does the negative sign matter?

A negative total energy does not mean the satellite has negative kinetic energy.

It means the satellite is gravitationally bound to Earth. It does not have enough mechanical energy to escape to infinite separation while retaining a positive speed.

Think of an overdrawn bank account. Moving from a balance of minus 100 dollars to minus 60 dollars is an increase of 40 dollars, even though the balance remains negative.

Orbital energy behaves similarly. A higher circular orbit has a less negative total energy.

For a bound circular orbit:

Notice that the total energy is negative because the magnitude of the gravitational potential energy is twice the kinetic energy.

02The formula for energy required to raise a circular orbit

Suppose a satellite moves from an initial circular orbit of radius \(r_i\) to a final circular orbit of radius \(r_f\).

Its initial total energy is:

\[ E_i=-\frac{GMm}{2r_i} \]

Its final total energy is:

\[ E_f=-\frac{GMm}{2r_f} \]

The change in total mechanical energy is:

\[ \Delta E=E_f-E_i \]

Substituting:

\[ \Delta E= -\frac{GMm}{2r_f} +\frac{GMm}{2r_i} \]

Therefore:

\[ \boxed{ \Delta E= \frac{GMm}{2} \left( \frac{1}{r_i}-\frac{1}{r_f} \right) } \]

This is the main calculation formula.

If the satellite moves to a higher circular orbit, \(r_f>r_i\). Therefore, the expression in brackets is positive, and energy must be added.

If the satellite moves to a lower circular orbit, \(r_f<r_i\). The expression becomes negative, meaning the satellite’s total mechanical energy decreases.

An important distinction: orbital radius is not altitude

This is one of the easiest mistakes to make in an HSC calculation.

Orbital altitude is measured from Earth’s surface. Orbital radius is measured from Earth’s centre.

You must use orbital radius in the energy equation.

The relationship is:

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

where:

• \(r\) is orbital radius, in metres.
• \(R_E\) is Earth’s radius, in metres.
• \(h\) is orbital altitude, in metres.

For our calculations, use these illustrative Earth values:

You can combine \(G\) and Earth’s mass into \(GM_E\) to make calculations shorter.

For example, a satellite at an altitude of \(400\,\mathrm{km}\) has an orbital radius of:

\[ r=6.37\times10^6+400\times10^3 \]

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

It does not have an orbital radius of \(400\,\mathrm{km}\).

A reliable calculation method

When a question asks how much energy is required to raise a satellite’s circular orbit, follow this sequence:

1. Identify the satellite’s mass.
2. Convert both orbital altitudes into metres.
3. Add Earth’s radius to each altitude to obtain \(r_i\) and \(r_f\).
4. Substitute into the change-in-energy equation.
5. Check the sign of your answer and explain what it means.

Use:

\[ \Delta E= \frac{GM_E m}{2} \left( \frac{1}{r_i}-\frac{1}{r_f} \right) \]

The final answer must be in joules.

Do not use \(mgh\) for a large change in orbital altitude. That approximation assumes gravitational acceleration is approximately constant, which is not appropriate when a satellite moves substantial distances away from Earth.

03Worked examples

04What happens if the satellite moves into a lower orbit?

Consider the reverse of our first worked example.

An \(800\,\mathrm{kg}\) satellite moves from a circular orbit at \(1050\,\mathrm{km}\) altitude to a circular orbit at \(450\,\mathrm{km}\) altitude.

The initial and final radii have simply swapped.

Therefore:

\[ \Delta E=-1.89\times10^9\,\mathrm{J} \]

The negative result means the satellite loses \(1.89\,\mathrm{GJ}\) of total orbital mechanical energy.

But there is another apparent contradiction.

The satellite moves faster in its lower circular orbit, so its kinetic energy increases.

For this change:

\[ \Delta K=+1.89\times10^9\,\mathrm{J} \]

Meanwhile:

\[ \Delta U=-3.78\times10^9\,\mathrm{J} \]

The decrease in gravitational potential energy is larger than the increase in kinetic energy. Therefore, the total mechanical energy decreases.

This is why you should never use the satellite’s speed alone to decide whether its total energy has increased or decreased.

05What the energy calculation does not tell you

The equation:

\[ \Delta E= \frac{GM_E m}{2} \left( \frac{1}{r_i}-\frac{1}{r_f} \right) \]

compares two circular orbital states.

It tells you the net change in the satellite’s orbital mechanical energy. It does not describe every moment of the transfer between those orbits.

In a typical orbital transfer, the satellite uses engine burns to change its velocity and follows a non-circular path between the two circular orbits.

During an engine-off part of the journey, gravity transfers energy between kinetic and gravitational potential energy, but total mechanical energy stays constant.

Also, the calculated energy increase is not the same as the chemical energy stored in rocket propellant or the electrical energy consumed by a real propulsion system. It is the ideal net mechanical energy gained by the satellite, assuming no energy losses and a constant satellite mass.

For a more detailed explanation of the two-burn transfer process, see Energy Changes When Satellites Change Orbit: HSC Physics.

For a fuller derivation of the circular-orbit energy relationship, see Total Energy in Circular Orbits for HSC Physics.

06Questions and solutions

The following questions build from a standard orbit-raising calculation towards more challenging applications of energy conservation. Use:

\[ R_E=6.37\times10^6\,\mathrm{m} \]

\[ GM_E=3.99\times10^{14}\,\mathrm{m^3\,s^{-2}} \]

Unless stated otherwise, assume a spherical Earth, negligible satellite mass compared with Earth’s mass, constant satellite mass, and no atmospheric drag or other external perturbations.

Question 1

A \(500\,\mathrm{kg}\) satellite moves from a circular orbit \(400\,\mathrm{km}\) above Earth’s surface to a circular orbit \(800\,\mathrm{km}\) above Earth’s surface.

Calculate the increase in its total orbital mechanical energy.

Solution 1

The satellite requires an increase of approximately \(8.22\times10^8\,\mathrm{J}\) in total orbital mechanical energy.

First, convert both altitudes into orbital radii.

\[ \begin{aligned} r_i&=6.37\times10^6+0.40\times10^6\\ &=6.77\times10^6\,\mathrm{m} \end{aligned} \]

\[ \begin{aligned} r_f&=6.37\times10^6+0.80\times10^6\\ &=7.17\times10^6\,\mathrm{m} \end{aligned} \]

Apply the energy-change equation:

\[ \Delta E= \frac{GM_E m}{2} \left( \frac{1}{r_i}-\frac{1}{r_f} \right) \]

Substitute:

\[ \begin{aligned} \Delta E &=\frac{(3.99\times10^{14})(500)}{2}\\ &\quad\times \left( \frac{1}{6.77\times10^6} -\frac{1}{7.17\times10^6} \right)\\ &=(9.975\times10^{16})(8.240\times10^{-9})\\ &=8.22\times10^8\,\mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{\Delta E=8.22\times10^8\,\mathrm{J}} \]

The answer is positive because the final circular orbit has a greater radius and a less negative total mechanical energy.

Question 2

Two identical \(500\,\mathrm{kg}\) satellites are to be raised by the same increase in altitude, \(400\,\mathrm{km}\).

• Satellite A moves from \(400\,\mathrm{km}\) to \(800\,\mathrm{km}\) altitude.
• Satellite B moves from \(20\,000\,\mathrm{km}\) to \(20\,400\,\mathrm{km}\) altitude.

Calculate the energy required for each satellite.

Determine the ratio of the energy required for A to the energy required for B.

Explain why the two energy requirements are different, despite both satellites gaining the same altitude.

Solution 2

Satellite A requires approximately \(14.5\) times as much energy as satellite B.

The key is that equal changes in altitude do not produce equal changes in orbital energy. The gravitational field becomes weaker as orbital radius increases.

For satellite A:

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

\[ r_{f,A}=7.17\times10^6\,\mathrm{m} \]

Using the equation:

\[ \Delta E_A= \frac{GM_E m}{2} \left( \frac{1}{r_{i,A}}-\frac{1}{r_{f,A}} \right) \]

\[ \boxed{\Delta E_A=8.22\times10^8\,\mathrm{J}} \]

For satellite B:

\[ r_{i,B}=6.37\times10^6+20.00\times10^6 \]

\[ r_{i,B}=26.37\times10^6\,\mathrm{m} \]

Similarly:

\[ r_{f,B}=26.77\times10^6\,\mathrm{m} \]

Substitute:

\[ \begin{aligned} \Delta E_B &=\frac{(3.99\times10^{14})(500)}{2}\\ &\quad\times \left( \frac{1}{26.37\times10^6} -\frac{1}{26.77\times10^6} \right)\\ &=(9.975\times10^{16})(5.666\times10^{-10})\\ &=5.65\times10^7\,\mathrm{J} \end{aligned} \]

Now calculate the ratio:

\[ \frac{\Delta E_A}{\Delta E_B} = \frac{8.22\times10^8}{5.65\times10^7} \]

\[ \boxed{\frac{\Delta E_A}{\Delta E_B}\approx14.5} \]

Satellite A needs approximately 14.5 times more energy.

Why?

The energy equation can also be written as:

\[ \Delta E= \frac{GM_E m}{2} \frac{r_f-r_i}{r_i r_f} \]

For the same satellite mass and the same increase in radius, the product \(r_i r_f\) determines the difference.

Satellite B starts much farther from Earth. Its gravitational potential energy changes less for the same increase in orbital radius.

The tempting assumption is that equal altitude increases require equal energy increases. They do not, because the gravitational field is not uniform.

Question 3

A \(900\,\mathrm{kg}\) satellite moves from a circular orbit of radius \(7.00\times10^6\,\mathrm{m}\) to another circular orbit of radius \(1.10\times10^7\,\mathrm{m}\).

A student makes the following statement:

“The propulsion system must supply the entire increase in gravitational potential energy. The satellite’s kinetic energy remains constant because it is still in orbit.”

Calculate:

(a) The change in total mechanical energy.

(b) The change in kinetic energy.

(c) The change in gravitational potential energy.

(d) Whether the student’s statement is correct.

Solution 3

The total mechanical energy increases by \(9.33\times10^9\,\mathrm{J}\). The student’s statement is incorrect because the satellite’s kinetic energy decreases as it moves into a higher circular orbit.

(a) Change in total mechanical energy

Use:

\[ \Delta E= \frac{GM_E m}{2} \left( \frac{1}{r_i}-\frac{1}{r_f} \right) \]

Substitute:

\[ \begin{aligned} \Delta E &=\frac{(3.99\times10^{14})(900)}{2}\\ &\quad\times \left( \frac{1}{7.00\times10^6} -\frac{1}{1.10\times10^7} \right)\\ &=(1.7955\times10^{17})(5.195\times10^{-8})\\ &=9.33\times10^9\,\mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{\Delta E=+9.33\times10^9\,\mathrm{J}} \]

(b) Change in kinetic energy

For circular orbits:

\[ K=\frac{GM_E m}{2r} \]

Therefore:

\[ \Delta K=K_f-K_i \]

Comparing this with the total energy equation gives:

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

Hence:

\[ \boxed{\Delta K=-9.33\times10^9\,\mathrm{J}} \]

The satellite loses kinetic energy because its final circular orbital speed is lower.

(c) Change in gravitational potential energy

For circular orbital endpoints:

\[ U=-\frac{GM_E m}{r} \]

Therefore:

\[ \Delta U=U_f-U_i \]

Comparing this with the total energy equation:

\[ \boxed{\Delta U=2\Delta E} \]

Hence:

\[ \boxed{\Delta U=+1.87\times10^{10}\,\mathrm{J}} \]

(d) Assess the student’s statement

The student is incorrect.

The gravitational potential energy increases by approximately \(1.87\times10^{10}\,\mathrm{J}\), but the kinetic energy decreases by approximately \(9.33\times10^9\,\mathrm{J}\).

Adding these changes:

\[ \begin{aligned} \Delta E&=\Delta K+\Delta U\\ &=-9.33\times10^9+1.87\times10^{10}\\ &=9.33\times10^9\,\mathrm{J} \end{aligned} \]

The propulsion system must supply the net increase, not the entire increase in gravitational potential energy.

The important condition is that both the initial and final orbits are circular. That is what allows us to use the relationship \(\Delta K=-\Delta E\).

Question 4

A \(700\,\mathrm{kg}\) satellite is travelling along a non-circular transfer orbit.

At a particular instant:

• Its distance from Earth’s centre is \(1.20\times10^7\,\mathrm{m}\).
• Its speed is \(4.50\times10^3\,\mathrm{m\,s^{-1}}\).

Calculate its total mechanical energy at this instant.

Determine whether the satellite is gravitationally bound to Earth.

A student attempts to use:

\[ E=-\frac{GM_E m}{2r} \]

Explain why this is or is not appropriate.

Solution 4

The satellite’s total mechanical energy is approximately \(-1.62\times10^{10}\,\mathrm{J}\), so it is gravitationally bound to Earth.

However, the student’s use of the circular-orbit equation is incorrect because the satellite is travelling along a non-circular path.

For a satellite at any position and speed, use the general energy equation:

\[ E=K+U \]

Therefore:

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

Step 1: Calculate kinetic energy.

\[ K=\frac12mv^2 \]

\[ \begin{aligned} K&=\frac12(700)(4.50\times10^3)^2\\ &=7.09\times10^9\,\mathrm{J} \end{aligned} \]

Step 2: Calculate gravitational potential energy.

\[ U=-\frac{GM_E m}{r} \]

\[ \begin{aligned} U&=-\frac{(3.99\times10^{14})(700)} {1.20\times10^7}\\ &=-2.33\times10^{10}\,\mathrm{J} \end{aligned} \]

Step 3: Calculate total mechanical energy.

\[ \begin{aligned} E&=K+U\\ &=7.09\times10^9-2.33\times10^{10}\\ &=-1.62\times10^{10}\,\mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{E=-1.62\times10^{10}\,\mathrm{J}} \]

The negative total energy means the satellite is gravitationally bound to Earth under the stated model.

Step 4: Explain why the circular-orbit shortcut is invalid.

For a circular orbit, the required speed at a given radius is:

\[ v_c=\sqrt{\frac{GM_E}{r}} \]

Substituting:

\[ \begin{aligned} v_c&=\sqrt{\frac{3.99\times10^{14}}{1.20\times10^7}}\\ &=5.77\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

But the satellite’s actual speed is only:

\[ v=4.50\times10^3\,\mathrm{m\,s^{-1}} \]

Its speed does not satisfy the circular-orbit condition.

Therefore, it is not in a circular orbit, and the shortcut:

\[ E=-\frac{GM_E m}{2r} \]

cannot be used.

This question highlights an important limitation: knowing a satellite’s distance from Earth is not enough to calculate its total energy unless you also know its speed or know that it is in a circular orbit.

Question 5

A \(1000\,\mathrm{kg}\) satellite is initially in a circular orbit with radius:

\[ r_i=7.00\times10^6\,\mathrm{m} \]

Its intended final orbit is circular, with radius:

\[ r_f=2r_i \]

An engineer calculates the required increase in total orbital mechanical energy correctly.

However, instead of using a two-stage transfer, the engineer applies one instantaneous prograde tangential burn at the initial orbital radius. The burn adds exactly the energy calculated for the initial and final circular states.

The satellite then travels under gravity alone. When it first reaches \(r=2r_i\), it is still moving outwards.

Answer the following:

(a) Calculate the required increase in total orbital mechanical energy.

(b) Calculate the satellite’s speed when it first reaches \(r=2r_i\).

(c) Has the satellite entered the intended final circular orbit? Explain why or why not.

Assume an ideal instantaneous burn, no atmospheric drag, and a constant satellite mass.

Solution 5

The required energy increase is \(1.43\times10^{10}\,\mathrm{J}\). However, the satellite is not in the intended circular orbit when it first reaches \(2r_i\), even though its speed magnitude equals the circular orbital speed at that radius.

This is the key distinction: having the correct total energy and speed magnitude does not necessarily mean having the correct velocity direction for a circular orbit.

(a) Calculate the required energy increase

Start with:

\[ \Delta E= \frac{GM_E m}{2} \left( \frac{1}{r_i}-\frac{1}{r_f} \right) \]

Since:

\[ r_f=2r_i \]

we can substitute:

\[ \Delta E= \frac{GM_E m}{2} \left( \frac{1}{r_i}-\frac{1}{2r_i} \right) \]

Simplifying:

\[ \Delta E=\frac{GM_E m}{4r_i} \]

Substitute the values:

\[ \begin{aligned} \Delta E &=\frac{(3.99\times10^{14})(1000)} {4(7.00\times10^6)}\\ &=\frac{3.99\times10^{17}}{2.80\times10^7}\\ &=1.425\times10^{10}\,\mathrm{J} \end{aligned} \]

Therefore:

\[ \boxed{\Delta E=1.43\times10^{10}\,\mathrm{J}} \]

(b) Calculate the speed at the target radius

The burn adds exactly the energy required to change between the two circular orbital energy states.

Therefore, immediately after the burn, the satellite’s total mechanical energy is equal to the energy of the intended final circular orbit:

\[ E=-\frac{GM_E m}{2r_f} \]

Since \(r_f=2r_i\):

\[ E=-\frac{GM_E m}{4r_i} \]

Substituting:

\[ E=-1.425\times10^{10}\,\mathrm{J} \]

During the engine-off journey, total mechanical energy remains constant.

When the satellite reaches \(r=2r_i\), its gravitational potential energy is:

\[ U=-\frac{GM_E m}{2r_i} \]

Substituting:

\[ \begin{aligned} U&=-\frac{(3.99\times10^{14})(1000)} {2(7.00\times10^6)}\\ &=-2.85\times10^{10}\,\mathrm{J} \end{aligned} \]

Using:

\[ E=K+U \]

Rearrange:

\[ K=E-U \]

\[ \begin{aligned} K&=-1.425\times10^{10} -(-2.85\times10^{10})\\ &=1.425\times10^{10}\,\mathrm{J} \end{aligned} \]

Now calculate speed:

\[ K=\frac12mv^2 \]

\[ v=\sqrt{\frac{2K}{m}} \]

\[ \begin{aligned} v&=\sqrt{\frac{2(1.425\times10^{10})}{1000}}\\ &=\sqrt{2.85\times10^7}\\ &=5.34\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

Therefore:

\[ \boxed{v=5.34\times10^3\,\mathrm{m\,s^{-1}}} \]

(c) Determine whether the satellite is in the intended circular orbit

No. The satellite is not yet in the intended circular orbit.

This is where the tempting argument becomes interesting.

The circular orbital speed at \(r_f=2r_i\) is:

\[ v_c=\sqrt{\frac{GM_E}{r_f}} \]

Substitute:

\[ \begin{aligned} v_c&=\sqrt{\frac{3.99\times10^{14}} {1.40\times10^7}}\\ &=5.34\times10^3\,\mathrm{m\,s^{-1}} \end{aligned} \]

This is exactly the speed magnitude calculated in part (b).

A student might therefore conclude that the satellite has entered the intended circular orbit.

But there is a missing condition.

For a circular orbit, the satellite’s velocity must be entirely tangential to the orbit. Its radial velocity must be zero.

At the specified instant, the satellite is still moving outwards. Therefore, its velocity has a non-zero radial component.

Its total speed has the correct magnitude, but some of that speed is directed radially outwards. Its tangential component is consequently smaller than the speed required for a circular orbit.

The satellite is therefore still following a non-circular trajectory.

In an ideal instantaneous-impulse model, a second manoeuvre could redirect its velocity so that it becomes purely tangential, without changing its speed magnitude. This would change its angular momentum while leaving its kinetic energy and total mechanical energy unchanged.

That is the deeper point: total mechanical energy determines how much energy the satellite has, but it does not, by itself, determine the shape or direction of its orbit.

The original energy calculation was completely correct. What failed was the assumption that supplying the correct energy in a single burn automatically produces the desired circular orbit.

07Where to go next

Once you can calculate the energy difference between two circular orbits, the next step is understanding how a satellite actually gets between them.

A two-burn orbital transfer introduces another important distinction: the satellite gains energy during engine burns, but its total mechanical energy remains constant while it coasts under gravity.

That connection is developed in Energy Changes When Satellites Change Orbit: HSC Physics.