How to Find Kinetic Energy from Potential Energy in a Circular Orbit

A satellite has gravitational potential energy \(U=-8.0\times10^8\ \mathrm{J}\), measured relative to zero at infinity. If it is travelling in a circular orbit, would its kinetic energy be \(8.0\times10^8\ \mathrm{J}\), or half that amount?

It is half: its kinetic energy is \(K=+4.0\times10^8\ \mathrm{J}\).

For a circular orbit, the relationship is:

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

The negative sign matters. Gravitational potential energy is negative, while kinetic energy is always positive. More importantly, this relationship comes from the circular-orbit condition. It does not apply to every object that is gravitationally bound.

Let’s derive the relationship, understand what happens to the total mechanical energy, and work through the calculations you’ll need to recognise in HSC Physics questions.

01Why does kinetic energy equal half the magnitude of gravitational potential energy?

The important piece of information is that the satellite is in a circular orbit.

A satellite travelling in a circle constantly changes the direction of its velocity. This means it has a centripetal acceleration directed towards the centre of the circle.

What provides this acceleration?

Gravity.

The gravitational force between the planet and satellite acts towards the planet’s centre. It supplies the centripetal force required to maintain the circular orbit.

This gives us the starting equation.

Step 1: Equate gravitational force and centripetal force

Consider a planet with mass \(M\) and a satellite with mass \(m\), travelling in a circular orbit of radius \(r\).

The gravitational force is:

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

The centripetal force required for circular motion is:

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

Here:

• \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
• \(M\) is the mass of the planet, in kilograms.
• \(m\) is the mass of the satellite, in kilograms.
• \(r\) is the distance between the centres of the two objects, in metres.
• \(v\) is the orbital speed, in metres per second.

Since gravity supplies the centripetal force:

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

Notice that the satellite mass \(m\) appears on both sides. We can cancel it.

Rearranging gives:

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

This equation tells us the orbital speed required for a circular orbit at radius \(r\).

Step 2: Substitute into the kinetic energy equation

The kinetic energy of the satellite is:

\[
K=\frac{1}{2}mv^2
\]

We have just established that:

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

Substituting:

\[
K=\frac{1}{2}m\left(\frac{GM}{r}\right)
\]

Therefore:

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

Step 3: Compare this with gravitational potential energy

When gravitational potential energy is defined as zero at infinity, the gravitational potential energy of the satellite is:

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

Compare this with the kinetic energy equation:

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

The only difference is the negative sign and the factor of two.

Therefore:

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

Or, using the magnitude of gravitational potential energy:

\[
\boxed{K=\frac{|U|}{2}}
\]

That’s the relationship you need.

You don’t have to calculate the satellite’s speed, gravitational force, or orbital radius if the question already gives you its gravitational potential energy and confirms that the orbit is circular.

02Understand the three different energies

A common mistake is confusing kinetic energy, gravitational potential energy, and total mechanical energy.

They are related, but they are not interchangeable.

Let’s define each one carefully.

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

\[
E=K+U
\]

Since \(K=-U/2\):

\[
E=-\frac{U}{2}+U
\]

Therefore:

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

This gives us three useful relationships for circular orbits:

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

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

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

What does bound energy actually mean?

A gravitationally bound satellite has negative total mechanical energy:

\[
E<0
\]

This means it does not have enough mechanical energy to escape to infinity without additional energy being supplied.

There is another useful term: binding energy.

Binding energy is the positive amount of energy required to take the satellite from its orbit to infinity, leaving it with zero final speed.

We can write:

\[
B=-E
\]

For a circular orbit, because \(E=-K\):

\[
\boxed{B=K}
\]

This is an important distinction.

The binding energy is positive, the total mechanical energy is negative, and the kinetic energy is positive.

For example, if a satellite has:

• \(U=-8.0\times10^8\ \mathrm{J}\)
• \(K=+4.0\times10^8\ \mathrm{J}\)
• \(E=-4.0\times10^8\ \mathrm{J}\)

Then \(4.0\times10^8\ \mathrm{J}\) of additional energy is required for it to reach infinity with zero final speed.

It is not necessary to supply the full magnitude of its gravitational potential energy.

For a more detailed explanation of the relationship between kinetic, potential, and total mechanical energy, see Total Energy in Circular Orbits for HSC Physics.

03How to check the relationship using orbital measurements

The relationship \(K=-U/2\) is a theoretical prediction from Newtonian gravity and circular motion, rather than a result that needs to be measured directly in a laboratory.

However, we can describe how observations of an actual satellite could test the model.

Imagine a tracking station on the planet’s surface exchanging timed radio signals with a satellite. Repeated measurements allow its position to be reconstructed over time.

For a circular orbit, the satellite travels one circumference in one orbital period.

The circumference is:

\[
C=2\pi r
\]

If the orbital period is \(T\), the measured orbital speed is:

\[
v=\frac{2\pi r}{T}
\]

Newtonian gravity independently predicts:

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

The two speeds should agree if the circular-orbit model and the assumptions used are appropriate.

If the satellite’s speed changes significantly as it moves around the planet, the circular-orbit assumption needs to be reconsidered.

The important lesson is that knowing the satellite is in a circular orbit gives us information that gravitational potential energy alone would not normally provide.

It fixes the orbital speed, allowing us to calculate kinetic energy.

04The important exception: not every bound orbit is circular

Here’s where students can make a convincing mistake.

Suppose a satellite is at a particular distance from a planet. You know its gravitational potential energy is negative, so you conclude that it must be in a circular orbit.

That conclusion is incorrect.

An elliptical orbit is also gravitationally bound. Its satellite can have negative total mechanical energy without satisfying \(K=-U/2\) at every point.

Let’s see why.

Consider a satellite at a particular orbital radius with circular orbital speed \(v_c\).

Its circular-orbit kinetic energy is:

\[
K_c=\frac{1}{2}mv_c^2
\]

Now suppose its speed is increased to \(1.2v_c\), while its position remains unchanged.

Its new kinetic energy is:

\[
K=\frac{1}{2}m(1.2v_c)^2
\]

Squaring the speed factor gives:

\[
K=1.44\left(\frac{1}{2}mv_c^2\right)
\]

Therefore:

\[
K=1.44K_c
\]

But we already know that:

\[
K_c=\frac{|U|}{2}
\]

So:

\[
K=1.44\left(\frac{|U|}{2}\right)
\]

\[
\boxed{K=0.72|U|}
\]

The kinetic energy is now 72% of the magnitude of gravitational potential energy, not 50%.

Is the satellite still bound?

Its total mechanical energy is:

\[
E=K+U
\]

Since \(U=-|U|\):

\[
E=0.72|U|-|U|
\]

\[
\boxed{E=-0.28|U|}
\]

Its total energy is still negative, so it remains gravitationally bound.

It is simply no longer in a circular orbit.

This comparison exposes an important distinction:

• Circular orbit: \(K=|U|/2\).
• Bound orbit: \(E<0\), but the kinetic energy need not equal half the magnitude of gravitational potential energy.
• Escape threshold: \(E=0\).

In fact, the escape speed at a given radius is:

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

Since circular orbital speed is:

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

We obtain:

\[
\boxed{v_{\mathrm{esc}}=\sqrt{2}v_c}
\]

A satellite travelling at \(1.2v_c\) is still below escape speed.

This is why you must check the actual orbital condition before using the half-energy relationship.

05A quick method for HSC exam questions

When you encounter a question involving gravitational potential energy and kinetic energy, identify what information the question gives you before choosing an equation.

A few details are worth checking before you substitute values:

1. Keep the sign of gravitational potential energy. If \(U=-6.0\times10^9\ \mathrm{J}\), use \(K=-U/2\), not \(K=U/2\).
2. Use the centre-to-centre orbital radius. If a satellite is at altitude \(h\) above a planet of radius \(R\), use \(r=R+h\).
3. Check that the orbit is circular. The half-energy relationship depends on this condition.
4. Keep kinetic energy and total energy separate. A negative total mechanical energy does not mean negative kinetic energy.

The following worked examples show how to apply these ideas.

06Worked examples

07Questions and solutions

Try each question before reading its solution. The questions progress from straightforward substitutions to problems where you need to decide whether the circular-orbit relationship is actually valid.

Question 1

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

\[
U=-1.6\times10^9\ \mathrm{J}
\]

Calculate:

a) Its kinetic energy.

b) Its total mechanical energy.

Solution 1

The kinetic energy is \(8.0\times10^8\ \mathrm{J}\), and its total mechanical energy is \(-8.0\times10^8\ \mathrm{J}\).

Because the satellite is in a circular orbit, we can use:

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

Part a: Kinetic energy

\[
K=-\frac{-1.6\times10^9}{2}
\]

\[
\boxed{K=8.0\times10^8\ \mathrm{J}}
\]

Part b: Total mechanical energy

Use:

\[
E=K+U
\]

Substitute:

\[
E=8.0\times10^8-1.6\times10^9
\]

\[
\boxed{E=-8.0\times10^8\ \mathrm{J}}
\]

The negative total energy indicates a gravitationally bound satellite.

The key distinction is that kinetic energy is positive, even though total mechanical energy is negative.

Question 2

A satellite of mass \(2.0\times10^3\ \mathrm{kg}\) travels in a circular orbit around a planet with mass \(5.0\times10^{24}\ \mathrm{kg}\).

Its orbital radius, measured from the planet’s centre, is \(1.0\times10^7\ \mathrm{m}\).

Use:

\[
G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}
\]

Calculate:

a) The gravitational potential energy.

b) The kinetic energy.

c) The total mechanical energy.

d) The orbital speed.

Solution 2

The gravitational potential energy is \(-6.67\times10^{10}\ \mathrm{J}\), the kinetic energy is \(3.34\times10^{10}\ \mathrm{J}\), the total mechanical energy is \(-3.34\times10^{10}\ \mathrm{J}\), and the orbital speed is approximately \(5.77\times10^3\ \mathrm{m\,s^{-1}}\).

Part a: Gravitational potential energy

Use:

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

Substitute:

\[
U=-\frac{
(6.67\times10^{-11})
(5.0\times10^{24})
(2.0\times10^3)
}{
1.0\times10^7
}
\]

Therefore:

\[
\boxed{U=-6.67\times10^{10}\ \mathrm{J}}
\]

Part b: Kinetic energy

The satellite is in a circular orbit, so:

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

\[
K=-\frac{-6.67\times10^{10}}{2}
\]

\[
\boxed{K=3.34\times10^{10}\ \mathrm{J}}
\]

Part c: Total mechanical energy

\[
E=K+U
\]

\[
E=3.335\times10^{10}
-6.67\times10^{10}
\]

\[
\boxed{E=-3.34\times10^{10}\ \mathrm{J}}
\]

The negative value confirms that the satellite is gravitationally bound.

Part d: Orbital speed

Use:

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

Substitute:

\[
v^2=
\frac{
(6.67\times10^{-11})(5.0\times10^{24})
}{
1.0\times10^7
}
\]

\[
v^2=3.335\times10^7
\]

Taking the square root:

\[
v=\sqrt{3.335\times10^7}
\]

\[
\boxed{v=5.77\times10^3\ \mathrm{m\,s^{-1}}}
\]

This is approximately \(5.77\ \mathrm{km\,s^{-1}}\).

An important observation is that the satellite’s mass affects its kinetic and gravitational potential energies, but not its circular orbital speed around the same planet at the same radius.

Question 3

Two satellites, A and B, travel in circular orbits around the same planet.

Satellite A has mass \(m\) and orbital radius \(r\).

Satellite B has mass \(3m\) and orbital radius \(2r\).

Determine:

a) The ratio of their orbital speeds, \(v_B/v_A\).

b) The ratio of their kinetic energies, \(K_B/K_A\).

c) The ratio of the magnitudes of their gravitational potential energies.

d) Which satellite has the more negative total mechanical energy?

Solution 3

Satellite A has the greater orbital speed, but satellite B has 1.5 times the kinetic energy and 1.5 times the magnitude of gravitational potential energy. Satellite B also has the more negative total mechanical energy.

The key is to account for both satellite mass and orbital radius.

Part a: Ratio of orbital speeds

For circular orbits:

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

Both satellites orbit the same planet, so \(GM\) is identical.

Therefore:

\[
\frac{v_B}{v_A}
=
\sqrt{\frac{r_A}{r_B}}
\]

Substitute the orbital radii:

\[
\frac{v_B}{v_A}
=
\sqrt{\frac{r}{2r}}
\]

\[
\boxed{\frac{v_B}{v_A}=\frac{1}{\sqrt{2}}}
\]

Satellite B travels more slowly because it is further from the planet.

Part b: Ratio of kinetic energies

Use:

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

For satellite A:

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

For satellite B:

\[
K_B=\frac{GM(3m)}{2(2r)}
\]

Therefore:

\[
\frac{K_B}{K_A}
=
\frac{3m/(4r)}{m/(2r)}
\]

\[
\boxed{\frac{K_B}{K_A}=\frac32}
\]

Although B travels more slowly, its mass is three times greater.

The increased mass more than compensates for the reduction in orbital speed.

Part c: Ratio of gravitational potential energy magnitudes

The magnitude of gravitational potential energy is:

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

For the two satellites:

\[
\frac{|U_B|}{|U_A|}
=
\frac{3m/(2r)}{m/r}
\]

\[
\boxed{\frac{|U_B|}{|U_A|}=\frac32}
\]

Satellite B has 1.5 times the magnitude of gravitational potential energy.

Remember that both potential energies are negative. Satellite B’s gravitational potential energy is therefore more negative.

Part d: Compare total mechanical energies

For a circular orbit:

\[
E=-K
\]

Therefore:

\[
\frac{E_B}{E_A}
=
\frac{K_B}{K_A}
\]

\[
\boxed{\frac{E_B}{E_A}=\frac32}
\]

Since both total energies are negative, this means:

\[
E_B=1.5E_A
\]

Satellite B has the more negative total mechanical energy.

The interesting result is that the satellite travelling more slowly has greater kinetic energy and a more negative total energy. Its greater mass changes the energy calculations even though it does not change the circular orbital speed at a given radius.

Question 4

A satellite is travelling under the influence of a planet’s gravitational field. At a particular point in its orbit, its gravitational potential energy is:

\[
U=-1.20\times10^9\ \mathrm{J}
\]

Its total mechanical energy is:

\[
E=-4.00\times10^8\ \mathrm{J}
\]

The satellite has non-zero angular momentum.

Determine:

a) Its kinetic energy at this point.

b) Whether the satellite is gravitationally bound.

c) Whether the satellite is in a circular orbit.

d) How its speed compares with the circular orbital speed at the same radius.

Solution 4

The satellite has kinetic energy \(8.00\times10^8\ \mathrm{J}\). It is gravitationally bound, but it is not in a circular orbit. Its speed is approximately 1.15 times the circular orbital speed at the same radius.

This question tests whether you can distinguish negative total energy from the circular-orbit condition.

Part a: Calculate kinetic energy

Start with:

\[
E=K+U
\]

Rearrange:

\[
K=E-U
\]

Substitute:

\[
K=-4.00\times10^8
-(-1.20\times10^9)
\]

\[
\boxed{K=8.00\times10^8\ \mathrm{J}}
\]

Notice that subtracting a negative gravitational potential energy increases the result.

Part b: Determine whether the satellite is bound

A satellite is gravitationally bound when its total mechanical energy is negative.

Here:

\[
E=-4.00\times10^8\ \mathrm{J}
\]

Therefore, it is gravitationally bound.

Part c: Determine whether the orbit is circular

For a circular orbit:

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

The kinetic energy required for a circular orbit at this radius would be:

\[
K_c=-\frac{-1.20\times10^9}{2}
\]

\[
K_c=6.00\times10^8\ \mathrm{J}
\]

However, the actual kinetic energy is:

\[
K=8.00\times10^8\ \mathrm{J}
\]

These values are different.

Therefore, the satellite is not in a circular orbit.

Because it is bound and has non-zero angular momentum, its ideal Newtonian trajectory is elliptical.

Part d: Compare orbital speeds

At the same radius, the circular orbital kinetic energy is \(6.00\times10^8\ \mathrm{J}\).

The actual kinetic energy is \(8.00\times10^8\ \mathrm{J}\).

Using \(K=\frac12mv^2\), the ratio of speeds is:

\[
\frac{v}{v_c}
=
\sqrt{\frac{K}{K_c}}
\]

Substitute:

\[
\frac{v}{v_c}
=
\sqrt{\frac{8.00\times10^8}{6.00\times10^8}}
\]

\[
\frac{v}{v_c}
=
\sqrt{\frac43}
\]

\[
\boxed{\frac{v}{v_c}\approx1.15}
\]

The satellite is travelling approximately 15% faster than the circular orbital speed at that radius.

The tempting mistake is to assume that negative total mechanical energy automatically means circular motion. It doesn’t. Negative energy tells you that the satellite is bound; the relationship \(K=-U/2\) requires the additional condition of a circular orbit.

Question 5

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

Its initial kinetic and gravitational potential energies are:

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

\[
U_i=-1.2\times10^{10}\ \mathrm{J}
\]

The satellite is eventually placed into a new circular orbit with radius \(2r\).

Assume the satellite’s mass and the planet’s mass remain constant. Ignore atmospheric drag, other bodies, and energy losses. The only non-gravitational work done on the satellite is supplied by its propulsion system.

Calculate:

a) The final kinetic energy.

b) The final gravitational potential energy.

c) The initial and final total mechanical energies.

d) The net work supplied by the propulsion system.

A student argues that the propulsion system must supply an amount of energy equal to the increase in gravitational potential energy.

Explain why this argument is incorrect.

Solution 5

The final kinetic energy is \(3.0\times10^9\ \mathrm{J}\), the final gravitational potential energy is \(-6.0\times10^9\ \mathrm{J}\), and the propulsion system supplies a net \(3.0\times10^9\ \mathrm{J}\) of work.

The student’s argument is incorrect because gravitational potential energy is not the only energy that changes. The satellite’s kinetic energy also decreases.

This is the important interaction between the two forms of energy.

Part a: Calculate final kinetic energy

For a circular orbit:

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

The satellite moves to an orbit with twice the original radius.

Therefore:

\[
K_f=\frac{GMm}{2(2r)}
\]

Comparing this with the initial kinetic energy:

\[
K_f=\frac{K_i}{2}
\]

Substitute:

\[
K_f=\frac{6.0\times10^9}{2}
\]

\[
\boxed{K_f=3.0\times10^9\ \mathrm{J}}
\]

The kinetic energy halves when the orbital radius doubles.

Part b: Calculate final gravitational potential energy

Gravitational potential energy is:

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

Doubling the orbital radius halves the magnitude of gravitational potential energy.

Therefore:

\[
U_f=\frac{U_i}{2}
\]

Substitute:

\[
U_f=\frac{-1.2\times10^{10}}{2}
\]

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

Notice that gravitational potential energy increases from \(-1.2\times10^{10}\ \mathrm{J}\) to \(-6.0\times10^9\ \mathrm{J}\).

It becomes less negative.

Part c: Calculate initial and final total mechanical energies

The initial total mechanical energy is:

\[
E_i=K_i+U_i
\]

\[
E_i=6.0\times10^9-1.2\times10^{10}
\]

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

The final total mechanical energy is:

\[
E_f=K_f+U_f
\]

\[
E_f=3.0\times10^9-6.0\times10^9
\]

\[
\boxed{E_f=-3.0\times10^9\ \mathrm{J}}
\]

The total mechanical energy has increased, becoming less negative.

Part d: Calculate the work supplied by propulsion

The net work done by the propulsion system equals the change in the satellite’s total mechanical energy, under the stated assumptions.

Therefore:

\[
W_{\mathrm{prop}}=\Delta E
\]

\[
W_{\mathrm{prop}}=E_f-E_i
\]

Substitute:

\[
W_{\mathrm{prop}}
=
-3.0\times10^9
-(-6.0\times10^9)
\]

\[
\boxed{W_{\mathrm{prop}}=+3.0\times10^9\ \mathrm{J}}
\]

The propulsion system supplies \(3.0\times10^9\ \mathrm{J}\) of net work.

Why is the student’s argument incorrect?

The student’s argument focuses only on the change in gravitational potential energy.

Let’s calculate both changes:

Change in gravitational potential energy:

\[
\Delta U=U_f-U_i
\]

\[
\Delta U=
-6.0\times10^9
-(-1.2\times10^{10})
\]

\[
\boxed{\Delta U=+6.0\times10^9\ \mathrm{J}}
\]

Change in kinetic energy:

\[
\Delta K=K_f-K_i
\]

\[
\Delta K=3.0\times10^9-6.0\times10^9
\]

\[
\boxed{\Delta K=-3.0\times10^9\ \mathrm{J}}
\]

Now combine them:

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

\[
\Delta E=-3.0\times10^9
+6.0\times10^9
\]

\[
\boxed{\Delta E=+3.0\times10^9\ \mathrm{J}}
\]

The increase in gravitational potential energy is partly offset by the decrease in kinetic energy.

This is why the net work supplied is only half the increase in gravitational potential energy.

The tempting approach is to calculate \(\Delta U\) and treat it as the total energy supplied. The correct approach is to calculate the change in total mechanical energy, including both kinetic and potential energy.

This is a particularly useful distinction when studying satellite orbit transfers.

08Where to go next

The key result to remember is:

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

But always check that the satellite is in a circular orbit before using it.

Once you can distinguish kinetic energy, gravitational potential energy, and total mechanical energy, you can tackle a closely related problem: calculating how much energy is required to move a satellite from one circular orbit to another.

The important step is to compare the initial and final total mechanical energies, rather than looking at gravitational potential energy alone.