How to Find Altitude from a Fraction of Surface Gravity

A satellite is at a point where the gravitational field strength is one-quarter of its value at the planet’s surface. How far above the surface is it?

The answer is one planet-radius above the surface, not two. The satellite is twice the planet’s radius from its centre, and doubling the centre-to-object distance reduces gravitational field strength to one-quarter.

There is a general formula for solving this type of question. If the gravitational field strength at altitude \(h\) is a fraction \(f\) of its surface value, then:

\[ \boxed{h=R\left(\frac{1}{\sqrt f}-1\right)} \]

Here, \(R\) is the planet’s radius, \(h\) is the altitude above its surface, and \(f\) is the fraction of surface gravitational field strength.

The important part is getting the distance right. Gravity depends on distance from the planet’s centre, not altitude above its surface.

01Understand the two different distances

Imagine a spherical planet with an object somewhere above its surface.

There are three distances to keep track of:

• \(R\): the planet’s radius, measured from its centre to its surface.
• \(h\): the object’s altitude, measured from the surface.
• \(r\): the object’s distance from the planet’s centre.

These distances are related by:

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

Why does this matter?

Newton’s law of universal gravitation uses the centre-to-centre distance between two objects. It does not use the altitude above the surface.

For example, suppose a planet has a radius of \(6\,000\ \mathrm{km}\).

An object at an altitude of \(6\,000\ \mathrm{km}\) is actually:

\[ r=6\,000+6\,000 \]

\[ r=12\,000\ \mathrm{km} \]

from the planet’s centre.

It is therefore at twice the planet’s radius, even though its altitude is only one planet-radius.

This is the distinction that makes the entire calculation work.

02Derive the altitude formula

The gravitational field strength outside a spherical planet is:

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

where:

• \(g\) is gravitational field strength in \(\mathrm{N\,kg^{-1}}\).
• \(G\) is the universal gravitational constant.
• \(M\) is the planet’s mass in kilograms.
• \(r\) is the distance from the planet’s centre in metres.

At the planet’s surface, the centre-to-object distance is simply \(R\).

Therefore, surface gravitational field strength is:

\[ g_s=\frac{GM}{R^2} \]

At altitude \(h\), the centre-to-object distance becomes \(R+h\), giving:

\[ g_h=\frac{GM}{(R+h)^2} \]

We want to know the altitude where gravitational field strength is a particular fraction of its surface value.

Define this fraction as:

\[ f=\frac{g_h}{g_s} \]

Substituting the two expressions:

\[ f= \frac{GM/(R+h)^2}{GM/R^2} \]

Notice that \(GM\) appears in both the numerator and denominator. It cancels.

This leaves:

\[ f=\frac{R^2}{(R+h)^2} \]

Taking the positive square root:

\[ \sqrt f=\frac{R}{R+h} \]

Rearranging to make \(h\) the subject:

\[ R+h=\frac{R}{\sqrt f} \]

Therefore:

\[ \boxed{h=R\left(\frac{1}{\sqrt f}-1\right)} \]

This is the equation to use whenever a question gives you a fraction of surface gravitational field strength and asks you to find the altitude.

What does the formula tell us?

Consider three different locations around the same planet.

Notice the pattern.

If the distance from the centre doubles, gravitational field strength becomes one-quarter.

If the distance from the centre triples, gravitational field strength becomes one-ninth.

But the altitude does not double or triple in the same way because the object’s starting position is already one radius away from the centre.

A useful prediction

Suppose someone tells you that an object is at an altitude of \(2R\).

Would you predict that its gravitational field strength is one-quarter of the surface value?

It is actually one-ninth.

The object is \(3R\) from the centre, so:

\[ \frac{g_h}{g_s} = \left(\frac{R}{3R}\right)^2 = \frac{1}{9} \]

The inverse-square relationship always uses \(r\), not \(h\).

03Don’t confuse gravitational field, force, and potential

A common source of mistakes is using the correct distance in the wrong equation.

These four quantities describe different aspects of gravity.

The key distinction is that gravitational field strength and force follow an inverse-square relationship, while gravitational potential and potential energy follow an inverse-distance relationship.

Consider an object at a distance of \(2R\) from the planet’s centre.

Its altitude is \(R\).

Gravitational field strength

\[ g=\frac{1}{4}g_s \]

Gravitational force

For the same object, its gravitational force is also one-quarter of its surface value.

This follows from \(F_g=mg\), because the object’s mass has not changed.

Gravitational potential

Taking gravitational potential to be zero at infinity:

\[ V=-\frac{GM}{r} \]

At \(2R\):

\[ V=-\frac{GM}{2R} \]

The potential is half its surface value, not one-quarter.

Gravitational potential energy

For an object of unchanged mass:

\[ U=mV \]

Its gravitational potential energy is also half its surface value.

Because gravitational potential and potential energy are negative under the zero-at-infinity convention, moving further away makes them less negative.

This distinction is important: a field-strength fraction of \(1/4\) does not mean a potential fraction of \(1/4\).

If you want to explore the relationship between gravitational potential and field strength further, see Can Gravitational Potential Be Zero When Gravity Isn’t?.

04How could we experimentally test the prediction?

The inverse-square equation predicts that gravitational field strength decreases with altitude.

But how could we check that prediction without directly measuring the gravitational force on a satellite?

A pendulum provides one way to investigate the change.

Setting up the experiment

Imagine a simple pendulum consisting of a small, dense bob suspended from a fixed pivot by a string of length \(L\).

A photogate is positioned at the bottom of the bob’s swing and connected to an electronic timer. As the bob passes through the photogate, the timer records its motion.

The experiment is performed at two locations:

• A station close to sea level.
• A mountain station \(1.00\ \mathrm{km}\) above sea level.

The same pendulum is used at both locations.

The investigator measures the time for 100 complete oscillations at each location, then divides the total time by 100 to obtain the period.

The period is the time taken for one complete oscillation.

For small oscillations, a simple pendulum follows:

\[ T=2\pi\sqrt{\frac{L}{g}} \]

where:

• \(T\) is the period in seconds.
• \(L\) is the pendulum length in metres.
• \(g\) is the local gravitational field strength.

What should the experiment show?

At the mountain station, gravitational field strength should be slightly weaker.

Because the pendulum’s period increases when \(g\) decreases, we predict that the pendulum will swing slightly more slowly at the higher altitude.

We can calculate the expected difference.

For the same pendulum length:

\[ \frac{T_h}{T_s} = \sqrt{\frac{g_s}{g_h}} \]

Using the inverse-square relationship:

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

Therefore:

\[ \frac{T_h}{T_s} = \frac{R+h}{R} \]

or:

\[ \boxed{\frac{T_h}{T_s}=1+\frac{h}{R}} \]

Take Earth’s radius as \(6.37\times10^6\ \mathrm m\), and use an altitude difference of \(1\,000\ \mathrm m\).

The predicted period ratio is:

\[ \begin{aligned} \frac{T_h}{T_s} &=1+\frac{1\,000}{6.37\times10^6}\\ &=1.000157 \end{aligned} \]

Suppose the pendulum has a period of \(2.00000\ \mathrm s\) at sea level.

At the higher station:

\[ T_h=2.00000\times1.000157 \]

\[ T_h=2.000314\ \mathrm s \]

The difference is only about \(0.000314\ \mathrm s\) per oscillation.

That seems tiny. However, over 100 oscillations, the predicted difference becomes approximately:

\[ 100(0.000314)=0.0314\ \mathrm s \]

An illustrative set of readings might therefore be:

These are illustrative values, not recorded experimental results.

The longer period at the higher station is consistent with the prediction that gravitational field strength decreases with altitude.

What could go wrong with the original design?

Suppose the investigator uses two different pendulums, one at each station.

If the second pendulum has a slightly longer string, its period will also increase. The investigator might incorrectly attribute this difference to weaker gravity.

The experiment has not isolated the variable being investigated.

A better design uses:

• The same pendulum at both locations.
• The same measured string length, with temperature changes controlled.
• Small oscillation angles.
• Electronic timing over many oscillations.
• Stations at approximately the same latitude, with corrections for local terrain, geology, and Earth’s rotation.

These controls help separate the effect of altitude from other influences on local gravitational acceleration.

The important limitation is that the experiment tests the ideal spherical-Earth prediction only after local variations have been considered. Real Earth measurements are more complicated than the simple model.

The mathematical model nevertheless gives a clear prediction: increasing the distance from Earth’s centre decreases gravitational field strength according to an inverse-square relationship.

05Worked examples

06Questions and solutions

The following questions progress from a straightforward altitude calculation to problems involving gravitational potential, force, energy, and circular orbits.

Question 1

A spherical planet has a radius of \(5.0\times10^6\ \mathrm m\).

An object is placed above the planet’s surface where the gravitational field strength is one-ninth of its surface value.

Calculate the object’s altitude above the planet’s surface.

Solution 1

The required altitude is:

\[ \boxed{h=1.0\times10^7\ \mathrm m} \]

The important step is recognising that one-ninth of the surface gravitational field strength occurs at three times the planet’s radius from its centre.

Start with:

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

Given:

\[ f=\frac19 \]

Taking the square root:

\[ \frac13=\frac{R}{R+h} \]

Rearranging:

\[ R+h=3R \]

Therefore:

\[ h=2R \]

Substituting the planet’s radius:

\[ h=2(5.0\times10^6) \]

\[ \boxed{h=1.0\times10^7\ \mathrm m} \]

The object’s distance from the centre is \(1.5\times10^7\ \mathrm m\), but its altitude is only \(1.0\times10^7\ \mathrm m\).

The distinction between these two distances is essential.

Question 2

A spherical planet has a radius of \(8.0\times10^6\ \mathrm m\).

Its gravitational potential at the surface is:

\[ V_s=-8.0\times10^7\ \mathrm{J\,kg^{-1}} \]

At a particular location above the surface, the gravitational potential is:

\[ V_h=-4.0\times10^7\ \mathrm{J\,kg^{-1}} \]

Assume gravitational potential is zero at infinity.

Calculate:

(a) The altitude of the location above the surface.

(b) The fraction of surface gravitational field strength at this location.

Explain why the potential fraction and field-strength fraction are different.

Solution 2

The altitude is \(8.0\times10^6\ \mathrm m\), and the gravitational field strength is one-quarter of its surface value.

The key is to use the inverse-distance relationship for potential rather than the inverse-square relationship for field strength.

Step 1: Compare the gravitational potentials.

Using:

\[ V=-\frac{GM}{r} \]

the ratio of potential at the location to potential at the surface is:

\[ \frac{V_h}{V_s}=\frac{R}{r} \]

Substituting:

\[ \frac{-4.0\times10^7}{-8.0\times10^7} = \frac{R}{r} \]

Therefore:

\[ 0.5=\frac{R}{r} \]

Rearranging:

\[ r=2R \]

Step 2: Calculate the altitude.

Remember:

\[ r=R+h \]

Therefore:

\[ h=2R-R \]

\[ h=R \]

Substituting:

\[ \boxed{h=8.0\times10^6\ \mathrm m} \]

Step 3: Calculate the gravitational field-strength fraction.

The gravitational field follows an inverse-square relationship:

\[ \frac{g_h}{g_s} = \left(\frac{R}{r}\right)^2 \]

Since \(r=2R\):

\[ \frac{g_h}{g_s} = \left(\frac{R}{2R}\right)^2 \]

\[ \boxed{\frac{g_h}{g_s}=\frac14} \]

The potential fraction is \(1/2\), while the field-strength fraction is \(1/4\).

Both results are correct because the two quantities have different relationships with distance.

A common mistake is to use the potential ratio directly as the field-strength ratio. The equations are not interchangeable.

Question 3

A spherical planet has a radius of \(4.0\times10^6\ \mathrm m\) and a surface gravitational field strength of \(15\ \mathrm{N\,kg^{-1}}\).

Two rovers are positioned above the surface:

• Rover A has a mass of \(30\ \mathrm{kg}\) and is at an altitude of \(R\).
• Rover B has a mass of \(60\ \mathrm{kg}\) and is at an altitude of \(2R\).

Calculate the gravitational field strength at each rover and determine which rover experiences the greater gravitational force.

Explain why the rover with the greater mass does not necessarily experience the greater gravitational force.

Solution 3

Rover A experiences the greater gravitational force, despite Rover B having twice its mass.

This happens because the gravitational field strength decreases with the square of the distance from the planet’s centre.

Step 1: Calculate the centre-to-rover distance for each rover.

For Rover A:

\[ r_A=R+R=2R \]

For Rover B:

\[ r_B=R+2R=3R \]

Step 2: Calculate the gravitational field strength at Rover A.

\[ g_A=g_s\left(\frac{R}{2R}\right)^2 \]

\[ g_A=15\times\frac14 \]

\[ \boxed{g_A=3.75\ \mathrm{N\,kg^{-1}}} \]

Step 3: Calculate the gravitational field strength at Rover B.

\[ g_B=g_s\left(\frac{R}{3R}\right)^2 \]

\[ g_B=15\times\frac19 \]

\[ \boxed{g_B=1.67\ \mathrm{N\,kg^{-1}}} \]

Step 4: Calculate the gravitational force on each rover.

Using:

\[ F_g=mg \]

For Rover A:

\[ F_A=30\times3.75 \]

\[ \boxed{F_A=112.5\ \mathrm N} \]

For Rover B:

\[ F_B=60\times1.67 \]

\[ \boxed{F_B\approx100\ \mathrm N} \]

Step 5: Compare the forces.

\[ \frac{F_A}{F_B} = \frac{112.5}{100} \]

\[ \boxed{\frac{F_A}{F_B}=1.125} \]

Rover A experiences approximately \(12.5\%\) more gravitational force than Rover B.

The tempting prediction is that Rover B must experience a greater force because it has twice the mass.

However, Rover B is also much further from the planet’s centre. Its gravitational field strength is only \(4/9\) of Rover A’s.

The reduction in field strength outweighs the increase in mass.

This is why gravitational field strength and gravitational force must be treated as separate quantities.

Question 4

A satellite of mass \(120\ \mathrm{kg}\) is slowly moved away from a spherical planet.

The planet has:

• Radius \(R=6.0\times10^6\ \mathrm m\).
• Surface gravitational field strength \(g_s=10\ \mathrm{N\,kg^{-1}}\).

The satellite moves from an initial altitude of \(R\) to a final altitude of \(2R\).

Assume the satellite moves slowly enough that its change in kinetic energy is negligible.

Calculate:

(a) The gravitational field strength at its initial position.

(b) The gravitational field strength at its final position.

(c) The change in gravitational potential energy.

(d) The external work required to move the satellite between these positions.

Solution 4

The gravitational field strength decreases from \(2.5\ \mathrm{N\,kg^{-1}}\) to approximately \(1.11\ \mathrm{N\,kg^{-1}}\).

The satellite’s gravitational potential energy increases by \(1.2\times10^9\ \mathrm J\), so the external work required is also \(1.2\times10^9\ \mathrm J\).

The important distinction is that gravitational field strength follows an inverse-square relationship, while gravitational potential energy follows an inverse-distance relationship.

Step 1: Calculate the initial centre-to-satellite distance.

\[ r_i=R+R=2R \]

Step 2: Calculate the initial gravitational field strength.

\[ g_i=g_s\left(\frac{R}{2R}\right)^2 \]

\[ g_i=\frac{10}{4} \]

\[ \boxed{g_i=2.5\ \mathrm{N\,kg^{-1}}} \]

Step 3: Calculate the final centre-to-satellite distance.

\[ r_f=R+2R=3R \]

Step 4: Calculate the final gravitational field strength.

\[ g_f=g_s\left(\frac{R}{3R}\right)^2 \]

\[ g_f=\frac{10}{9} \]

\[ \boxed{g_f=1.11\ \mathrm{N\,kg^{-1}}} \]

Step 5: Calculate the change in gravitational potential energy.

Gravitational potential energy is:

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

We can express \(GM\) using the surface gravitational field strength:

\[ g_s=\frac{GM}{R^2} \]

Therefore:

\[ GM=g_sR^2 \]

Substituting into the potential-energy equation:

\[ U=-\frac{mg_sR^2}{r} \]

At the initial position:

\[ U_i=-\frac{mg_sR}{2} \]

At the final position:

\[ U_f=-\frac{mg_sR}{3} \]

The change in potential energy is:

\[ \Delta U=U_f-U_i \]

Therefore:

\[ \Delta U = -\frac{mg_sR}{3} +\frac{mg_sR}{2} \]

Combining the fractions:

\[ \Delta U=\frac{mg_sR}{6} \]

Substituting the given values:

\[ \begin{aligned} \Delta U &=\frac{120\times10\times6.0\times10^6}{6}\\ &=1.2\times10^9\ \mathrm J \end{aligned} \]

Thus:

\[ \boxed{\Delta U=+1.2\times10^9\ \mathrm J} \]

The positive sign means the satellite’s gravitational potential energy has increased. Its energy has become less negative as it moves further away from the planet.

Step 6: Calculate the external work required.

Because the satellite moves slowly and its change in kinetic energy is negligible:

\[ W_{\text{external}}=\Delta U \]

Therefore:

\[ \boxed{W_{\text{external}}=1.2\times10^9\ \mathrm J} \]

Gravity does the opposite amount of work:

\[ W_{\text{gravity}}=-1.2\times10^9\ \mathrm J \]

A tempting shortcut would be to use \(W=mg_s\Delta h\), treating surface gravitational field strength as constant throughout the movement.

That would be incorrect because the gravitational field strength changes significantly as the satellite moves away from the planet.

For large changes in altitude, use gravitational potential energy rather than assuming a constant gravitational force.

Question 5

A satellite initially travels in a circular orbit around a spherical planet at an altitude of \(R\), where \(R\) is the planet’s radius.

Mission controllers want to move it into a higher circular orbit where the gravitational field strength is one-sixteenth of its surface value.

A trainee makes two predictions:

• The final altitude must be \(4R\), because the centre-to-satellite distance is four times the planet’s radius.
• The final orbital speed must be one-quarter of its initial orbital speed because gravitational field strength is one-sixteenth of its surface value.

Evaluate both predictions.

Calculate:

(a) The correct final altitude.

(b) The ratio of final orbital speed to initial orbital speed.

(c) The ratio of final orbital period to initial orbital period.

(d) Whether a second satellite with twice the mass would have a different orbital speed or period in the same final orbit.

Assume both satellites have negligible mass compared with the planet, and ignore atmospheric resistance and the gravitational influence of other bodies.

Solution 5

The trainee is wrong about both the final altitude and the orbital-speed ratio.

The correct final altitude is \(3R\), the final orbital speed is \(1/\sqrt{2}\) of the initial speed, and the final orbital period is \(2\sqrt{2}\) times the initial period.

The difficulty is that gravitational field strength, orbital speed, and orbital period depend on the orbital radius in different ways.

Step 1: Calculate the initial orbital radius.

The initial altitude is \(R\).

Therefore:

\[ r_i=R+R \]

\[ \boxed{r_i=2R} \]

This is the distance from the planet’s centre, not the altitude.

Step 2: Calculate the final orbital radius.

The final gravitational field strength is one-sixteenth of the surface value.

Using:

\[ \frac{g_f}{g_s} = \left(\frac{R}{r_f}\right)^2 \]

Substitute:

\[ \frac{1}{16} = \left(\frac{R}{r_f}\right)^2 \]

Taking the positive square root:

\[ \frac14=\frac{R}{r_f} \]

Therefore:

\[ \boxed{r_f=4R} \]

The trainee has correctly identified the final centre-to-satellite distance.

However, the question asks for altitude.

Step 3: Calculate the final altitude.

\[ r_f=R+h_f \]

Therefore:

\[ 4R=R+h_f \]

Rearranging:

\[ \boxed{h_f=3R} \]

The first prediction was incorrect because it confused the distance from the centre with the altitude above the surface.

Step 4: Calculate the orbital-speed ratio.

For a satellite in a circular orbit, gravity provides the centripetal force.

Therefore:

\[ F_g=F_c \]

Using the gravitational force equation and centripetal force equation:

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

Cancel the satellite’s mass and simplify:

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

Therefore:

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

For the same planet, \(GM\) is constant.

Consequently:

\[ \frac{v_f}{v_i} = \sqrt{\frac{r_i}{r_f}} \]

Substitute the two orbital radii:

\[ \begin{aligned} \frac{v_f}{v_i} &=\sqrt{\frac{2R}{4R}}\\ &=\sqrt{\frac12} \end{aligned} \]

Therefore:

\[ \boxed{\frac{v_f}{v_i}=\frac{1}{\sqrt2}\approx0.707} \]

The final orbital speed is approximately \(70.7\%\) of the initial orbital speed.

Why isn’t it one-quarter?

The trainee has assumed that orbital speed depends only on the square root of gravitational field strength.

But the circular orbital-speed equation can also be written as:

\[ v=\sqrt{gr} \]

The orbital radius changes as well as the gravitational field strength.

You cannot calculate the speed ratio by considering the change in \(g\) alone.

Step 5: Calculate the orbital-period ratio.

The orbital period is the time required to complete one full revolution.

For a circular orbit:

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

For satellites orbiting the same planet:

\[ T\propto r^{3/2} \]

Therefore:

\[ \frac{T_f}{T_i} = \left(\frac{r_f}{r_i}\right)^{3/2} \]

Substituting:

\[ \begin{aligned} \frac{T_f}{T_i} &=\left(\frac{4R}{2R}\right)^{3/2}\\ &=2^{3/2}\\ &=2\sqrt2 \end{aligned} \]

Thus:

\[ \boxed{\frac{T_f}{T_i}=2\sqrt2\approx2.83} \]

The satellite takes approximately \(2.83\) times as long to complete one orbit in its higher orbit.

This makes physical sense. It is travelling more slowly, and its circular path is also larger.

Step 6: Consider a second satellite with twice the mass.

The second satellite experiences twice the gravitational force at the same orbital radius:

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

However, its inertia also doubles.

When calculating its orbital speed:

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

the satellite’s mass cancels.

Therefore, at the same orbital radius, both satellites have the same circular orbital speed.

They also have the same orbital period because:

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

contains no satellite mass.

The second satellite experiences a greater gravitational force, but it does not need a different orbital speed or period.

This is the important connection between gravitational field strength and circular motion: gravity determines the centripetal acceleration, but the resulting orbital speed depends on both the strength of the field and the orbital radius.

For further practice with this relationship, see Gravity as Centripetal Force in Circular Orbits for HSC Physics.

The next time a question gives you a fraction of surface gravitational field strength, first calculate the distance from the planet’s centre. Once you have that distance, you can correctly determine altitude, gravitational force, potential energy, and even orbital motion.