How to Calculate a Satellite’s Centripetal Acceleration

A satellite 400 km above Earth is still accelerating towards Earth’s centre at about \(8.7\,\mathrm{m\,s^{-2}}\). For a circular orbit, calculate its centripetal acceleration using

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

The important detail is that \(r\) is the distance from the centre of Earth, not the satellite’s altitude above the surface. If the satellite is 400 km above Earth, its orbital radius is Earth’s radius plus 400 km.

Would you have expected the acceleration to be close to the familiar \(9.8\,\mathrm{m\,s^{-2}}\) at Earth’s surface? It should be. At 400 km altitude, the satellite is still relatively close to Earth’s centre, so gravity has only decreased modestly.

01Why the formula works

The calculation comes from combining Newton’s law of universal gravitation with the centripetal acceleration formula.

Imagine a satellite of mass \(m\) travelling in a circular orbit around a planet of mass \(M\).

Earth, or whichever central body is being considered, gravitationally attracts the satellite towards its centre. This gravitational force continually changes the direction of the satellite’s velocity, producing the acceleration needed to keep it moving in a circle.

For an ideal circular orbit, gravity supplies the entire centripetal force.

Newton’s law of universal gravitation gives:

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

Here:

• \(F_g\) is the gravitational force, measured in newtons (N).
• \(G\) is the universal gravitational constant.
• \(M\) is the mass of the central body, in kilograms.
• \(m\) is the satellite’s mass, in kilograms.
• \(r\) is the distance between the centres of the two bodies, in metres.

Newton’s second law tells us that the net force producing the satellite’s centripetal acceleration is:

\[ F_c=ma_c \]

Since gravity supplies this force, equate the expressions:

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

Cancel the satellite’s mass:

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

This is the main formula to remember when calculating the centripetal acceleration of a satellite in a circular orbit.

Notice something important: the satellite’s mass has disappeared.

A 200 kg satellite and a 2,000 kg satellite travelling in circular orbits at the same radius around the same planet have the same centripetal acceleration. The heavier satellite experiences a greater gravitational force, but it also has proportionally greater inertia.

What does the negative sign mean?

The equation above gives the magnitude of the acceleration, which is a positive number.

However, acceleration is a vector. It has a direction, and that direction is towards the centre of the orbit.

If we define the outward radial direction as positive, we can write:

\[ a_r=-\frac{GM}{r^2} \]

The negative sign indicates that the acceleration points inwards.

This is why centripetal acceleration is sometimes called radial acceleration. If you want to understand the direction in more detail, see Centripetal Acceleration: Why It Points Inward in HSC Physics.

Assumptions behind the calculation

The formula assumes:

• The satellite follows a circular orbit.
• The central body is spherical and can be treated as a point mass for the calculation.
• The satellite’s mass is negligible compared with the central body’s mass.
• Gravity is the only force acting on the satellite.
• Effects such as atmospheric drag and gravitational disturbances from other bodies are ignored.

These assumptions matter. In particular, the formula \(a_c=v^2/r\) describes circular motion. If a satellite is no longer travelling in a circular orbit, you cannot automatically use that expression as its actual acceleration.

02Where does the gravitational constant come from?

The value of \(G\) is usually supplied in HSC Physics questions:

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

But how was this number established?

The answer involves one of the most important gravitational experiments in physics: Henry Cavendish’s torsion balance experiment in 1798.

At the time, scientists understood that objects attracted each other gravitationally. However, measuring the tiny gravitational attraction between ordinary objects in a laboratory was extremely difficult.

Cavendish’s experiment was designed to make that tiny force measurable.

Cavendish’s torsion balance

The apparatus consisted of a light horizontal rod suspended from a thin vertical wire.

A small lead sphere was attached to each end of the rod. Two much larger lead spheres could be positioned close to the smaller ones.

The arrangement allowed the gravitational attraction between the large and small spheres to produce a tiny turning effect, or torque, on the suspended rod.

What did Cavendish expect to observe?

When the large lead spheres were brought close to the smaller spheres, their gravitational attraction should twist the suspended rod.

Moving the large spheres to opposite positions should reverse the direction of the torque, causing the rod to twist in the opposite direction.

The tiny movement could be measured using a telescope and a scale.

The experiment had a serious problem: the gravitational forces were extremely weak. Even a small air current could disturb the suspended rod and overwhelm the effect being measured.

Cavendish improved the experimental arrangement to reduce these disturbances. He placed the apparatus in a closed room, moved the large masses using a pulley system operated from outside, and observed the apparatus through a telescope. The room was also kept dark to reduce unwanted temperature differences.

The revised arrangement allowed the small gravitational deflection to be measured more reliably.

Cavendish reported the mean density of Earth, rather than presenting his result as a measurement of \(G\) in the modern form. Later physicists expressed the result in terms of the universal gravitational constant.

The historical distinction is worth remembering: Cavendish measured the attraction between laboratory masses and used it to determine Earth’s density. Modern physics describes the experiment as an early measurement of \(G\).

The historical details are discussed in the American Physical Society’s account of Cavendish’s experiment and this review of gravitational constant measurements.

How small was the gravitational force?

Consider an illustrative pair of lead spheres:

• Large sphere mass: \(158\,\mathrm{kg}\)
• Small sphere mass: \(0.73\,\mathrm{kg}\)
• Distance between their centres: \(0.23\,\mathrm{m}\)

Using Newton’s law:

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

Substituting:

\[ F_g= \frac{(6.67\times10^{-11})(158)(0.73)} {(0.23)^2} \]

\[ F_g\approx1.45\times10^{-7}\,\mathrm{N} \]

This is an extremely small force: approximately \(0.000000145\,\mathrm{N}\).

These are illustrative rounded values, not a reconstruction of Cavendish’s complete experimental measurements.

The experiment demonstrates why sensitive apparatus was necessary. It also provides the experimental background to the gravitational constant used in satellite calculations.

The same gravitational relationship applies whether the interacting masses are lead spheres in a laboratory or a planet and a satellite separated by thousands of kilometres.

03Choosing the right formula

You can calculate centripetal acceleration using different equations, depending on the information supplied.

All three expressions below are equivalent for a circular orbit.

The third expression comes from the relationship between orbital speed and period.

One complete revolution covers a circumference of \(2\pi r\). Therefore:

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

Substituting this into \(a_c=v^2/r\):

\[ a_c=\frac{1}{r} \left(\frac{2\pi r}{T}\right)^2 \]

Simplifying:

\[ \boxed{a_c=\frac{4\pi^2r}{T^2}} \]

Here:

• \(v\) is orbital speed in metres per second.
• \(T\) is orbital period in seconds.
• \(r\) is orbital radius in metres.

One important conversion: altitude to orbital radius

Questions frequently give the satellite’s altitude rather than its orbital radius.

If a satellite is at altitude \(h\) above a planet’s surface, use:

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

where:

• \(R\) is the radius of the planet.
• \(h\) is the satellite’s altitude.
• \(r\) is the distance from the planet’s centre.

For example, if Earth’s radius is \(6.37\times10^6\,\mathrm{m}\) and a satellite is 400 km above the surface:

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

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

Convert kilometres into metres before substituting into the equations.

If the question supplies gravitational acceleration at the planet’s surface, \(g_0\), you can also use:

\[ a_c=g_0\left(\frac{R}{r}\right)^2 \]

This follows from applying the inverse-square relationship between gravitational acceleration and distance from the planet’s centre.

04Worked examples

05Questions and solutions

The following questions progress from standard HSC calculations to more demanding problems involving multiple orbital relationships.

Question 1

A 500 kg satellite moves in a circular orbit 500 km above Earth’s surface.

Use the following information:

• Earth’s mass: \(5.97\times10^{24}\,\mathrm{kg}\)
• Earth’s radius: \(6.37\times10^6\,\mathrm{m}\)
• Gravitational constant: \(6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\)

Calculate the magnitude and direction of the satellite’s centripetal acceleration.

Solution 1

The satellite’s centripetal acceleration is \(8.44\,\mathrm{m\,s^{-2}}\), directed towards Earth’s centre.

Step 1: Calculate the orbital radius.

Convert the altitude:

\[ h=500\times10^3 =5.00\times10^5\,\mathrm{m} \]

Then:

\[ r=R_E+h \]

\[ r=6.37\times10^6+5.00\times10^5 \]

\[ r=6.87\times10^6\,\mathrm{m} \]

Step 2: Apply the gravitational acceleration formula.

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

Substitute:

\[ a_c= \frac{(6.67\times10^{-11})(5.97\times10^{24})} {(6.87\times10^6)^2} \]

\[ a_c= \frac{3.98\times10^{14}} {4.72\times10^{13}} \]

\[ \boxed{a_c=8.44\,\mathrm{m\,s^{-2}}} \]

Step 3: State the direction.

Gravitational acceleration always points towards the centre of the central body.

Therefore, the answer is \(8.44\,\mathrm{m\,s^{-2}}\) towards Earth’s centre.

Notice that the satellite’s mass of 500 kg was not needed. Its mass cancels when gravitational force is equated to centripetal force.

Question 2

An unmanned relay satellite travels in a circular orbit around an illustrative airless moon.

The following measurements are recorded:

• Moon radius: \(1.60\times10^6\,\mathrm{m}\)
• Satellite altitude: \(2.00\times10^5\,\mathrm{m}\)
• Orbital period: \(2.50\,\mathrm{h}\)

Calculate:

a) The satellite’s centripetal acceleration.

b) Its orbital speed.

Solution 2

The satellite’s centripetal acceleration is \(0.877\,\mathrm{m\,s^{-2}}\), and its orbital speed is approximately \(1.26\times10^3\,\mathrm{m\,s^{-1}}\).

Step 1: Calculate the orbital radius.

The orbital radius includes both the moon’s radius and the satellite’s altitude.

\[ r=R+h \]

\[ r=1.60\times10^6+2.00\times10^5 \]

\[ r=1.80\times10^6\,\mathrm{m} \]

Step 2: Convert the period into seconds.

\[ T=2.50\times3600 \]

\[ T=9000\,\mathrm{s} \]

Step 3: Calculate centripetal acceleration.

We have orbital radius and period, so:

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

Substitute:

\[ a_c= \frac{4\pi^2(1.80\times10^6)} {(9000)^2} \]

\[ a_c=\frac{71.06\times10^6}{81.0\times10^6} \]

\[ \boxed{a_c=0.877\,\mathrm{m\,s^{-2}}} \]

Step 4: Calculate orbital speed.

One complete orbit covers a distance of \(2\pi r\).

Therefore:

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

Substitute:

\[ v=\frac{2\pi(1.80\times10^6)}{9000} \]

\[ \boxed{v=1.26\times10^3\,\mathrm{m\,s^{-1}}} \]

The satellite travels at approximately \(1,260\,\mathrm{m\,s^{-1}}\), while its velocity continually changes direction.

Its acceleration points towards the moon’s centre.

Question 3

Two satellites, A and B, orbit the same spherical planet in separate circular orbits.

Their properties are:

Here, \(R\) is the radius of the planet, and both satellites orbit the same central mass.

Calculate:

a) The ratio of their centripetal accelerations, \(a_B/a_A\).

b) The ratio of their gravitational forces, \(F_B/F_A\).

c) The ratio of their orbital periods, \(T_B/T_A\).

Explain why the satellite with the greater mass does not necessarily have the greater centripetal acceleration.

Solution 3

The results are:

\[ \boxed{\frac{a_B}{a_A}=\frac49} \]

\[ \boxed{\frac{F_B}{F_A}=\frac43} \]

\[ \boxed{\frac{T_B}{T_A}\approx1.84} \]

Satellite B has a smaller centripetal acceleration, despite experiencing a greater gravitational force.

Step 1: Calculate the acceleration ratio.

For circular orbits around the same central mass:

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

Therefore:

\[ \frac{a_B}{a_A} = \frac{GM/(3R)^2}{GM/(2R)^2} \]

The common terms \(G\) and \(M\) cancel:

\[ \frac{a_B}{a_A} = \frac{(2R)^2}{(3R)^2} \]

\[ \boxed{\frac{a_B}{a_A}=\frac49} \]

Satellite B has only four-ninths the centripetal acceleration of satellite A.

Step 2: Calculate the gravitational force ratio.

Newton’s second law gives:

\[ F=ma \]

Therefore:

\[ \frac{F_B}{F_A} = \frac{m_Ba_B}{m_Aa_A} \]

Substitute the mass and acceleration ratios:

\[ \frac{F_B}{F_A} = \frac{1800}{600}\times\frac49 \]

\[ =3\times\frac49 \]

\[ \boxed{\frac{F_B}{F_A}=\frac43} \]

Satellite B experiences a gravitational force that is four-thirds that experienced by satellite A.

Step 3: Calculate the orbital period ratio.

For a circular orbit:

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

Combining this with \(a_c=GM/r^2\) gives:

\[ T=2\pi\sqrt{\frac{r^3}{GM}} \]

Since both satellites orbit the same planet, \(G\) and \(M\) cancel when taking the ratio.

\[ \frac{T_B}{T_A} = \left(\frac{r_B}{r_A}\right)^{3/2} \]

Substitute:

\[ \frac{T_B}{T_A} = \left(\frac{3R}{2R}\right)^{3/2} \]

\[ =\left(\frac32\right)^{3/2} \]

\[ \boxed{\frac{T_B}{T_A}\approx1.84} \]

Satellite B takes approximately 1.84 times as long to complete one orbit.

Why is the heavier satellite’s acceleration smaller?

The tempting prediction is that satellite B should have a greater acceleration because it experiences a greater gravitational force.

However, acceleration depends on both force and mass:

\[ a=\frac{F}{m} \]

Satellite B has three times the mass but is in an orbit with a larger radius. Its gravitational acceleration is reduced by the inverse-square relationship.

The increase in gravitational force does not outweigh the increase in satellite mass when determining acceleration.

Question 4

An Earth-observation satellite is reported to be in a circular orbit 700 km above Earth’s surface.

A tracking station reports that its orbital period is 95.0 minutes.

Use:

• Earth’s mass: \(5.97\times10^{24}\,\mathrm{kg}\)
• Earth’s radius: \(6.37\times10^6\,\mathrm{m}\)
• Gravitational constant: \(6.67\times10^{-11}\,\mathrm{N\,m^2\,kg^{-2}}\)

Assume the reported measurements are exact.

a) Calculate the centripetal acceleration predicted by Earth’s gravitational field.

b) Calculate the acceleration implied by the reported orbital period.

c) Determine whether the two measurements are consistent with a circular orbit. If they are not, calculate the orbital period that would be expected at the stated altitude.

Solution 4

The two measurements are inconsistent. The gravitational model predicts an acceleration of \(7.97\,\mathrm{m\,s^{-2}}\) and an orbital period of approximately 98.7 minutes. The reported period of 95.0 minutes implies a larger acceleration of \(8.59\,\mathrm{m\,s^{-2}}\).

Step 1: Calculate the orbital radius.

Convert the altitude:

\[ h=700\times10^3 =7.00\times10^5\,\mathrm{m} \]

Then:

\[ r=R_E+h \]

\[ r=6.37\times10^6+7.00\times10^5 \]

\[ r=7.07\times10^6\,\mathrm{m} \]

Step 2: Calculate acceleration using gravity.

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

Substitute:

\[ a_g= \frac{(6.67\times10^{-11})(5.97\times10^{24})} {(7.07\times10^6)^2} \]

\[ \boxed{a_g=7.97\,\mathrm{m\,s^{-2}}} \]

This is the acceleration predicted by Earth’s gravitational attraction.

Step 3: Calculate acceleration using the reported period.

Convert the reported period into seconds:

\[ T=95.0\times60 \]

\[ T=5700\,\mathrm{s} \]

Use:

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

Substitute:

\[ a_c= \frac{4\pi^2(7.07\times10^6)} {(5700)^2} \]

\[ \boxed{a_c=8.59\,\mathrm{m\,s^{-2}}} \]

Step 4: Determine the expected orbital period.

For a circular orbit, the gravitational acceleration must equal the centripetal acceleration.

Therefore:

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

Rearrange:

\[ T=2\pi\sqrt{\frac{r^3}{GM}} \]

Substitute:

\[ T= 2\pi\sqrt{ \frac{(7.07\times10^6)^3} {(6.67\times10^{-11})(5.97\times10^{24})} } \]

\[ T\approx5919\,\mathrm{s} \]

Convert to minutes:

\[ T=\frac{5919}{60} \]

\[ \boxed{T\approx98.7\,\mathrm{min}} \]

The expected period is approximately 98.7 minutes, not 95.0 minutes.

What does this tell us?

A satellite cannot simultaneously satisfy both calculated accelerations while maintaining the stated circular orbit under the given gravitational model.

At least one of the reported measurements or assumptions must be incorrect.

This is an important checking technique: two different methods should produce the same centripetal acceleration when the data describe the same circular orbit.

Question 5

Two satellites, A and B, travel in the same circular orbit around a spherical planet of mass \(M\).

Their orbital radius is \(r\), and their masses are:

• Satellite A: \(m\)
• Satellite B: \(4m\)

Initially, both satellites travel at the same circular orbital speed, \(v_c\).

A short engine burn then gives satellite B a tangential increase in speed. Its new speed becomes \(1.20v_c\), while its position remains unchanged during the brief impulse. The engine then switches off.

Assume:

• The planet’s mass is much greater than either satellite’s mass.
• No other forces act after the engine switches off.
• The satellites remain at the same instantaneous orbital radius immediately after the impulse.

Answer the following:

a) Before the engine burn, compare the gravitational forces acting on A and B.

b) Before the engine burn, compare their centripetal accelerations.

c) Immediately after the engine switches off, determine the gravitational acceleration of B relative to A.

d) A student claims that B’s centripetal acceleration immediately becomes \(1.44\) times its original value because \(a_c=v^2/r\). Explain whether this conclusion is correct.

e) Can B remain in the same circular orbit after the engine burn? Explain the physical reason.

Solution 5

Satellite B experiences four times the gravitational force on A, but its gravitational acceleration remains equal to A’s. After the speed increase, B cannot remain in the same circular orbit.

The key is to distinguish the gravitational acceleration actually acting on the satellite from the centripetal acceleration that would be required to maintain a particular circular path.

Step 1: Compare the gravitational forces before the engine burn.

For satellite A:

\[ F_{g,A}=\frac{GMm}{r^2} \]

For satellite B:

\[ F_{g,B}=\frac{GM(4m)}{r^2} \]

Therefore:

\[ F_{g,B}=4\frac{GMm}{r^2} \]

So:

\[ \boxed{F_{g,B}=4F_{g,A}} \]

Satellite B experiences four times the gravitational force because its mass is four times greater.

Step 2: Compare their initial accelerations.

Using Newton’s second law:

\[ a_A=\frac{F_{g,A}}{m} \]

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

For satellite B:

\[ a_B=\frac{F_{g,B}}{4m} \]

Substitute its gravitational force:

\[ a_B= \frac{4GMm/r^2}{4m} \]

Cancel the common factors:

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

Therefore:

\[ \boxed{a_A=a_B} \]

Both satellites have the same centripetal acceleration before the engine burn.

Step 3: Determine B’s gravitational acceleration after the engine burn.

The engine changes B’s velocity, but it does not change:

• The mass of the planet.
• The satellite’s instantaneous position.
• The gravitational field at that position.

The gravitational force on B therefore remains:

\[ F_{g,B}=\frac{4GMm}{r^2} \]

Its mass is still \(4m\), so:

\[ a_{g,B}=\frac{F_{g,B}}{4m} \]

\[ \boxed{a_{g,B}=\frac{GM}{r^2}} \]

Its gravitational acceleration has not increased.

Step 4: Investigate the student’s claim.

The student’s calculation is:

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

The original circular orbital speed satisfies:

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

After the engine burn, B’s speed is:

\[ v_B=1.20v_c \]

If we substitute this speed into the circular-motion expression:

\[ \frac{v_B^2}{r} = \frac{(1.20v_c)^2}{r} \]

\[ =\frac{1.44v_c^2}{r} \]

Using \(v_c^2=GM/r\):

\[ \frac{v_B^2}{r} = 1.44\frac{GM}{r^2} \]

The student’s numerical substitution is correct, but the interpretation is wrong.

The expression \(v^2/r\) gives the centripetal acceleration required to maintain circular motion at radius \(r\) with that speed.

It does not automatically give the actual acceleration when the satellite is no longer following that circular path.

Step 5: Decide whether B remains in its original circular orbit.

It cannot.

For a satellite to remain in a circular orbit at radius \(r\), its speed must satisfy:

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

The engine burn increases B’s speed to \(1.20v_c\).

However, its gravitational acceleration remains:

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

The increased speed would require a greater inward acceleration to maintain the original circular path:

\[ a_{\text{required}}=1.44\frac{GM}{r^2} \]

Gravity does not supply that greater acceleration.

The satellite therefore follows a different, non-circular path after the engine burn.

The important lesson is that \(v^2/r\) is not a formula to apply blindly whenever a satellite has a speed and a distance from a planet. It describes circular motion at that radius.

This is the distinction that separates a correct calculation from a correct physical explanation.

06What to calculate next

Once you can calculate centripetal acceleration at a known orbital radius, the next useful skill is reversing the process: using a satellite’s measured orbital period to determine its orbital radius.

That calculation uses the same relationship between gravitational attraction, orbital speed, and circular motion.

Continue with How to Find a Satellite’s Orbital Radius from Its Period.