Why Satellites Stay in Orbit Without a Supporting Surface

A satellite can travel around Earth for years without touching anything that supports it. A car going around a bend needs friction from the road. A ball spinning on a string needs tension from the string. So what keeps a satellite moving in a circle?

The answer is gravity. Earth's gravitational attraction provides the inward, or radial, force needed for circular motion. A satellite does not need a supporting surface because it is in continuous free fall: it is falling towards Earth while moving sideways fast enough to keep missing it.

This is not a special kind of force found only in space. It is the same gravitational force that makes a dropped object accelerate towards the ground. The difference is the satellite's horizontal speed and the curvature of its path.

## The key idea: gravity replaces the supporting force

Imagine a satellite travelling in a circular orbit around Earth. At every point in its orbit, its velocity is tangent to the circular path, while its acceleration points towards Earth's centre.

That inward acceleration continually changes the direction of the satellite's velocity. Its speed can remain constant even though its velocity is changing.

The force responsible is Earth's gravitational attraction. There is no normal reaction force because the satellite is not in contact with a surface.

[IMAGE: Earth with a satellite on a circular orbit, showing the satellite's tangential velocity arrow and an inward gravitational force arrow directed towards Earth's centre. Include a clear orbital radius from Earth's centre and distinguish the directions of velocity and force.]

The important distinction is between a force and the role that force plays.

*Centripetal force* means the net force directed towards the centre of a circular path. It is not an additional force that must be added to a free-body diagram.

For an orbiting satellite, gravity is the centripetal force. More precisely, gravity supplies the required net radial force.

Compare three familiar situations:

| Object moving in a circle | Real inward force | Does it need a supporting surface? |
|---|---|---|
| Car travelling around a flat bend | Friction between the tyres and road | Yes, the road also provides a normal reaction |
| Ball rotating on a horizontal string | Tension in the string | No supporting surface is needed for the horizontal circular motion |
| Satellite orbiting Earth | Earth's gravitational attraction | No |

The force changes between situations, but the requirement is the same: there must be a resultant force towards the centre of the circular path.

## Newton's cannonball: the historical idea behind an orbit

Why doesn't the Moon fall straight towards Earth?

Newton's explanation was that it does fall towards Earth. Its gravitational acceleration continually bends its path. However, the Moon also has a large sideways velocity, so it keeps moving forward as it falls.

This connected two situations that might initially seem unrelated: a projectile falling towards the ground and the Moon travelling around Earth.

Newton's 1687 work, *Principia Mathematica*, established the laws of motion and universal gravitation that allowed celestial motion to be explained using the same physics as motion on Earth. His famous cannonball thought experiment provides a particularly useful way to picture the argument. [Source: Smithsonian National Air and Space Museum and American Physical Society.] citeturn440139search0turn887944view3

### The thought experiment: a cannon on a mountain

Imagine a cannon positioned on top of a very high mountain. Its barrel points horizontally, in a direction tangent to Earth's surface.

Fire a cannonball.

At first, it travels forwards while gravity accelerates it downwards. Its path curves towards Earth, and eventually it hits the ground.

Now imagine firing another cannonball at a greater horizontal speed.

It travels further before reaching the ground. Give it even more speed, and it travels further still.

[IMAGE: Newton's cannonball thought experiment shown in cross-section. Draw a spherical Earth, a very high mountain with a horizontally pointing cannon, and several projectile paths at increasing launch speeds. Show the slow projectiles intersecting Earth's surface at progressively greater distances, and the special circular trajectory following Earth's curvature. Label the initial horizontal velocity and the inward gravitational direction.]

What happens at exactly the right speed?

The cannonball falls towards Earth, but Earth's surface curves away beneath it at the same rate. The ball continually misses the ground.

It is now in a circular orbit.

The crucial point is that gravity never stops acting. The cannonball is still accelerating towards Earth's centre. Its sideways velocity simply allows it to keep moving around Earth rather than colliding with the surface.

### What did the thought experiment actually establish?

It is important to distinguish a historical thought experiment from a physical experiment.

Newton did not collect measurements by firing a cannonball into orbit. The cannon was an idealised model used to predict what should happen under his laws of motion and gravity.

The reasoning produces three possible outcomes:

| Launch condition | Predicted motion |
|---|---|
| Insufficient horizontal speed | The projectile intersects Earth's surface |
| Correct speed for a circular orbit | The projectile continually falls around Earth |
| Greater than circular orbital speed | The projectile follows a different trajectory |

The final case needs some care. A speed greater than circular orbital speed does not automatically mean the object escapes Earth. It can follow an elliptical orbit instead. Escape requires an even greater speed. This distinction is explored further in the extension question at the end.

The thought experiment also has limitations. It assumes no air resistance, a spherical Earth, and an idealised gravitational model. A real cannonball fired through the atmosphere would experience drag, lose mechanical energy, and fail to maintain the predicted orbit.

Modern satellites are instead accelerated into carefully chosen trajectories by rockets. Once established in an appropriate orbit, gravity supplies the inward force required for their motion. Real satellites can still experience atmospheric drag, particularly in low Earth orbit, so their orbits may gradually change. citeturn440139search2turn440139search3

## What force actually acts on an orbiting satellite?

Now translate the historical idea into an HSC free-body diagram.

Consider a satellite travelling in a circular orbit around Earth.

There are two important quantities:

- The gravitational force acting on the satellite.
- The net inward force required to produce its circular acceleration.

The first is a real interaction between Earth and the satellite. The second describes the resultant force required for circular motion.

### 1. Gravitational force

Newton's law of universal gravitation gives:

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

where:

- \(F_g\) is the gravitational force in newtons (N).
- \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}\).
- \(M\) is Earth's mass, approximately \(5.97\times10^{24}\text{ kg}\).
- \(m\) is the satellite's mass in kilograms.
- \(r\) is the distance from Earth's centre to the satellite, in metres.

Notice that \(r\) is measured from Earth's centre, not from its surface.

If the satellite is at an altitude \(h\) above Earth's surface, then:

\[
r=R_E+h
\]

where \(R_E\) is Earth's radius.

[IMAGE: Earth shown in cross-section with a satellite above its surface. Clearly distinguish Earth's radius RE, the satellite's altitude h above the surface, and the orbital radius r measured from Earth's centre. Include the gravitational force arrow directed from the satellite towards Earth's centre.]

### 2. The force required for circular motion

For an object travelling in a circle, the required net inward force is:

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

where:

- \(F_c\) is the magnitude of the net inward force in newtons.
- \(m\) is the satellite's mass in kilograms.
- \(v\) is its speed in metres per second.
- \(r\) is its orbital radius in metres.

For a satellite in an ideal circular orbit, gravity is the only force acting in the radial direction.

Therefore:

\[
F_g=F_c
\]

Substituting the two equations:

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

Cancel the satellite's mass:

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

Rearranging:

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

Therefore:

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

This is the speed required for a circular orbit at radius \(r\), under the idealised assumption that Earth's gravity is the only significant force.

Notice something interesting: the satellite's mass has cancelled.

A heavier satellite requires a greater gravitational force to produce its circular acceleration, but its greater mass also means it requires a proportionally greater force for the same acceleration.

The resulting orbital speed does not depend on the satellite's mass.

## Worked example: How much does gravity pull on a satellite?

An \(800\text{ kg}\) satellite is travelling in a circular orbit \(400\text{ km}\) above Earth's surface.

Calculate:

a. The gravitational acceleration at the satellite's altitude.

b. The gravitational force acting on the satellite.

c. The speed required for a circular orbit at this altitude.

Use:

- Earth's mass: \(M=5.97\times10^{24}\text{ kg}\).
- Earth's radius: \(R_E=6.37\times10^6\text{ m}\).
- Gravitational constant: \(G=6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}\).

[IMAGE: Earth with an 800 kg satellite in a circular orbit 400 km above the surface. Label Earth's radius as 6.37 × 10^6 m, altitude as 400 km, orbital radius from Earth's centre, and tangential velocity. Show gravity pointing towards Earth's centre.]

**Step 1**: Convert the altitude into metres and calculate the orbital radius.

The orbital radius is measured from Earth's centre.

\[
h=400\text{ km}=4.00\times10^5\text{ m}
\]

Therefore:

\[
r=R_E+h
\]

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

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

**Step 2**: Calculate the gravitational acceleration.

From Newton's law of gravitation and \(F=ma\):

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

Substitute the values:

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

\[
g\approx8.69\text{ m s}^{-2}
\]

The gravitational acceleration is approximately \(8.69\text{ m s}^{-2}\), directed towards Earth's centre.

**Step 3**: Calculate the gravitational force.

\[
F_g=mg
\]

\[
F_g=800(8.69)
\]

\[
F_g=6.95\times10^3\text{ N}
\]

The satellite experiences approximately \(6,950\text{ N}\) of gravitational force.

**Step 4**: Calculate the required circular orbital speed.

Because gravity supplies the centripetal force:

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

Rearranging:

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

Substitute:

\[
v=
\sqrt{
\frac{(6.67\times10^{-11})(5.97\times10^{24})}
{6.77\times10^6}
}
\]

\[
v\approx7.67\times10^3\text{ m s}^{-1}
\]

The required circular orbital speed is approximately:

\[
\boxed{v=7.67\text{ km s}^{-1}}
\]

The result makes physical sense. Earth's gravitational attraction provides the inward force needed to continuously bend the satellite's path. No supporting surface is involved.

## The misconception: surely an astronaut in orbit has no weight?

This is where the idea of free fall becomes particularly useful.

Imagine an astronaut inside a spacecraft travelling in a circular orbit. The astronaut appears to float inside the cabin.

A common explanation is that gravity must be zero because the astronaut is weightless.

That explanation is incorrect.

At an altitude of \(400\text{ km}\), our previous example showed that gravitational acceleration is still approximately \(8.69\text{ m s}^{-2}\). This is close to the gravitational acceleration at Earth's surface.

So why does the astronaut float?

Because the astronaut and spacecraft are both accelerating towards Earth under gravity.

The astronaut does not need the spacecraft floor to push upwards to produce the required acceleration. Gravity already provides that acceleration.

The spacecraft and astronaut fall together.

[IMAGE: Side-by-side comparison. On the left, a person standing on Earth's surface with gravitational force downward and normal reaction upward. On the right, an astronaut floating inside a spacecraft in circular orbit, with Earth below and both spacecraft and astronaut following the same orbital path. Make the difference between surface contact and orbital free fall visually clear.]

### Gravitational weight versus apparent weight

There are two ideas worth separating.

**Gravitational weight** is the gravitational force acting on an object:

\[
W=mg
\]

**Apparent weight** is the support force experienced through contact with a surface, such as the normal reaction from a floor.

On Earth's surface, a stationary person has approximately balanced forces:

\[
N=mg
\]

where \(N\) is the normal reaction force.

In ideal orbital free fall, the person and spacecraft accelerate together. The floor does not need to provide an upward normal force to keep the person moving with the spacecraft.

Therefore:

\[
N=0
\]

The astronaut's gravitational weight is not zero. Their apparent weight is zero.

This is why an astronaut can experience apparent weightlessness while still being strongly affected by Earth's gravity. NASA describes this condition as free fall: the spacecraft and its occupants fall together under gravity. citeturn440139search7

## A contrasting case: a satellite versus a car around a bend

Let's make the distinction even clearer.

A car travels around a flat circular bend.

Its velocity points along the road, but its acceleration points towards the centre of the bend. The tyres need friction from the road to provide the inward force.

If there were no friction, the car would not follow the circular path.

Now consider a satellite.

Its velocity is also tangent to its circular path. Its acceleration is directed towards the centre of Earth. However, there are no tyres, no road, and no supporting surface.

Gravity provides the inward force instead.

| Feature | Car on a flat bend | Satellite in circular orbit |
|---|---|---|
| Tangential velocity | Along the road | Along the orbital path |
| Inward acceleration | Towards bend centre | Towards Earth's centre |
| Real inward force | Friction | Gravity |
| Normal reaction | Present | No contact reaction from Earth |
| Can the object maintain circular motion without its inward force? | No | No |

[IMAGE: Two scientifically accurate side-by-side free-body diagrams. Left: a car travelling around a flat bend, with friction towards the centre of the bend, normal reaction upwards, and gravity downwards. Right: a satellite orbiting Earth, with only an inward gravitational force arrow. Show the velocity tangent to each circular path. Do not add an outward centrifugal force to either inertial-frame diagram.]

The useful exam rule is simple:

**Identify the real forces first. Then calculate their resultant towards the centre.**

Never add a separate centripetal-force arrow on top of gravity, friction, tension, or magnetic force.

If you want to see how a different real force produces circular motion, compare this situation with [Circular Motion of a Charge in a Magnetic Field: HSC Physics](https://tutorgum.com/learn/resources/hsc-physics-circular-motion-of-a-charge-in-a-magnetic-field/).

## How to approach a satellite circular-motion question

When you see a satellite question in an HSC Physics exam, work through these decisions in order.

1. Identify the centre of the circular orbit. For a satellite orbiting Earth, this is Earth's centre.
2. Calculate the orbital radius using \(r=R_E+h\), if the altitude is provided.
3. Identify the real forces acting on the satellite. For an ideal circular orbit, gravity supplies the inward force.
4. Apply Newton's second law in the radial direction:

\[
\sum F_{\text{inward}}=\frac{mv^2}{r}
\]

5. Substitute the relevant gravitational or other force equation.
6. Check that your result has sensible units and represents the correct physical quantity.

If you need to revise the relationship between orbital speed, radius, and period, see [Speed, Radius and Period in Uniform Circular Motion](https://tutorgum.com/learn/resources/hsc-physics-speed-radius-and-period-in-circular-motion/).

## Questions and solutions

### Question 1

A \(600\text{ kg}\) satellite travels in a circular orbit around Earth at a radius of \(7.0\times10^6\text{ m}\) from Earth's centre.

Given:

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

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

Calculate:

a. The gravitational force acting on the satellite.

b. The speed required to maintain its circular orbit.

c. Explain why no normal reaction force is required.

[IMAGE: Earth shown as a circle with a 600 kg satellite on a circular orbit. Label the orbital radius as 7.0 × 10^6 m, and leave force and velocity arrows absent so students can identify and draw them.]

### Solution 1

The gravitational force is approximately \(4.88\times10^3\text{ N}\), and the required orbital speed is approximately \(7.54\times10^3\text{ m s}^{-1}\). Gravity supplies the inward force, so no normal reaction from a supporting surface is required.

**Part a: Gravitational force**

Use Newton's law of universal gravitation:

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

Substitute:

\[
F_g=
\frac{
(6.67\times10^{-11})
(5.97\times10^{24})(600)
}{
(7.0\times10^6)^2
}
\]

\[
F_g\approx4.88\times10^3\text{ N}
\]

**Part b: Orbital speed**

Gravity provides the required centripetal force:

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

Rearrange:

\[
v=\sqrt{\frac{F_gr}{m}}
\]

Substitute:

\[
v=
\sqrt{
\frac{(4.88\times10^3)(7.0\times10^6)}
{600}
}
\]

\[
v\approx7.54\times10^3\text{ m s}^{-1}
\]

**Part c: Normal reaction**

The satellite is not in contact with Earth's surface. Gravity is already providing the inward acceleration required for its circular motion.

Therefore, there is no normal reaction force from Earth.

The common mistake is to assume that every circular path needs a physical object pushing the moving object towards its centre. It does not. The required inward force can come from gravity, as it does here.

### Question 2

An astronaut with a mass of \(70\text{ kg}\) is inside a spacecraft travelling in a circular orbit \(300\text{ km}\) above Earth's surface.

Assume:

- Earth's radius is \(6.37\times10^6\text{ m}\).
- Earth's mass is \(5.97\times10^{24}\text{ kg}\).
- \(G=6.67\times10^{-11}\text{ N m}^2\text{ kg}^{-2}\).

a. Calculate the gravitational force acting on the astronaut.

b. Determine the normal reaction force acting on the astronaut in ideal orbital free fall.

c. Explain why the astronaut appears weightless even though Earth's gravitational force is still acting.

[IMAGE: A spacecraft in circular orbit around Earth, shown in cross-section. Include an astronaut standing beside a spring scale or floating just above the cabin floor. Label the 300 km altitude and leave the force arrows off the astronaut so students must construct the free-body diagram.]

### Solution 2

The astronaut experiences approximately \(627\text{ N}\) of gravitational force, but the normal reaction force is zero in ideal free fall. The astronaut appears weightless because the spacecraft and astronaut accelerate together under gravity.

**Part a: Gravitational force**

First calculate the orbital radius:

\[
r=R_E+h
\]

\[
r=6.37\times10^6+3.00\times10^5
\]

\[
r=6.67\times10^6\text{ m}
\]

Now calculate gravitational acceleration:

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

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

\[
g\approx8.95\text{ m s}^{-2}
\]

The gravitational force is:

\[
F_g=mg
\]

\[
F_g=70(8.95)
\]

\[
F_g\approx627\text{ N}
\]

**Part b: Normal reaction force**

The astronaut is in ideal free fall with the spacecraft.

Gravity provides the inward acceleration for both, so the spacecraft floor does not need to exert a normal reaction force to maintain the astronaut's motion.

Therefore:

\[
N=0\text{ N}
\]

**Part c: Apparent weightlessness**

The astronaut is still being pulled towards Earth by a gravitational force of approximately \(627\text{ N}\).

However, the astronaut and spacecraft experience the same gravitational acceleration. There is no supporting force pressing the astronaut against the floor.

An ordinary scale would therefore read zero in this idealised situation.

The trap is confusing the absence of a normal reaction force with the absence of gravity. They are not the same thing.

### Question 3

Two satellites, A and B, travel in circular orbits at the same radius of \(7.0\times10^6\text{ m}\) from Earth's centre.

Satellite A has a mass of \(300\text{ kg}\), while satellite B has a mass of \(900\text{ kg}\).

Assume both satellites are affected only by Earth's gravity.

a. Calculate the gravitational force acting on each satellite.

b. Determine the ratio of their gravitational accelerations.

c. Compare their required orbital speeds and periods.

d. A student claims that satellite B must orbit faster because it experiences a greater gravitational force. Explain the error.

[IMAGE: Two satellites of different sizes or masses shown at the same radial distance from Earth, following the same circular orbital path. Label A as 300 kg and B as 900 kg. Do not draw force arrows; leave students to determine the force and acceleration relationships.]

### Solution 3

Satellite B experiences three times the gravitational force of satellite A, but both have the same gravitational acceleration, orbital speed, and orbital period.

**Part a: Gravitational forces**

First calculate the gravitational acceleration at the shared orbital radius:

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

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

\[
g\approx8.13\text{ m s}^{-2}
\]

For satellite A:

\[
F_A=m_Ag
\]

\[
F_A=300(8.13)
\]

\[
F_A\approx2.44\times10^3\text{ N}
\]

For satellite B:

\[
F_B=m_Bg
\]

\[
F_B=900(8.13)
\]

\[
F_B\approx7.32\times10^3\text{ N}
\]

Therefore:

\[
\frac{F_B}{F_A}=3
\]

**Part b: Gravitational acceleration**

Using Newton's second law:

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

For both satellites, the gravitational acceleration is:

\[
a_A=a_B=8.13\text{ m s}^{-2}
\]

The greater force on satellite B is exactly matched by its greater mass.

**Part c: Orbital speed and period**

The orbital speed is:

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

Because both satellites have the same orbital radius around the same planet:

\[
v_A=v_B
\]

\[
v\approx7.54\times10^3\text{ m s}^{-1}
\]

The period is:

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

\[
T=
\frac{2\pi(7.0\times10^6)}
{7.54\times10^3}
\]

\[
T\approx5.83\times10^3\text{ s}
\]

Both satellites have a period of approximately \(97.2\text{ min}\).

**Part d: Explaining the misconception**

The student is confusing force with acceleration.

Satellite B experiences a greater gravitational force, but it also has three times the mass.

Newton's second law shows that:

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

The two effects cancel, producing the same gravitational acceleration.

The required circular orbital speed depends on the orbital radius and the mass of the planet, not the satellite's mass.

This is why a small satellite and a much heavier satellite can follow the same circular orbit at the same speed.

### Question 4

A satellite moves between two ideal circular orbits around Earth.

Initially, its altitude above Earth's surface is equal to Earth's radius:

\[
h_1=R_E
\]

It is then placed into a higher circular orbit with an altitude of:

\[
h_2=3R_E
\]

Assume Earth's mass and radius remain constant.

Determine the following ratios:

a. The new orbital radius compared with the original orbital radius.

b. The new gravitational force compared with the original gravitational force.

c. The new gravitational acceleration compared with the original gravitational acceleration.

d. The new orbital speed compared with the original orbital speed.

e. The new orbital period compared with the original orbital period.

Explain why using the altitude directly in the inverse-square law would produce an incorrect answer.

[IMAGE: Earth with two concentric circular satellite orbits. The lower orbit has altitude h1 = RE and orbital radius r1 = 2RE. The higher orbit has altitude h2 = 3RE and orbital radius r2 = 4RE. Label both distances from Earth's centre and both altitudes separately.]

### Solution 4

The new orbital radius is twice the original radius. Gravitational force and acceleration decrease to one-quarter, orbital speed decreases to \(1/\sqrt{2}\) of its original value, and the orbital period increases by a factor of \(2\sqrt{2}\).

**Part a: Orbital radius**

Remember that orbital radius is measured from Earth's centre.

Initially:

\[
r_1=R_E+h_1
\]

\[
r_1=R_E+R_E=2R_E
\]

In the new orbit:

\[
r_2=R_E+h_2
\]

\[
r_2=R_E+3R_E=4R_E
\]

Therefore:

\[
\frac{r_2}{r_1}
=\frac{4R_E}{2R_E}=2
\]

The orbital radius doubles.

**Part b: Gravitational force**

Newton's law of gravitation gives:

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

For the same satellite and planet:

\[
F_g\propto\frac{1}{r^2}
\]

Therefore:

\[
\frac{F_2}{F_1}
=
\left(\frac{r_1}{r_2}\right)^2
\]

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

\[
\frac{F_2}{F_1}=\frac14
\]

The gravitational force becomes one-quarter of its original value.

**Part c: Gravitational acceleration**

Since:

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

the same inverse-square relationship applies:

\[
\frac{g_2}{g_1}
=
\left(\frac{r_1}{r_2}\right)^2
\]

\[
\frac{g_2}{g_1}=\frac14
\]

The gravitational acceleration also becomes one-quarter of its original value.

**Part d: Orbital speed**

For a circular orbit:

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

Therefore:

\[
\frac{v_2}{v_1}
=
\sqrt{\frac{r_1}{r_2}}
\]

\[
=
\sqrt{\frac12}
\]

\[
\frac{v_2}{v_1}=\frac{1}{\sqrt2}
\]

The new orbital speed is approximately \(0.707\) times the original speed.

**Part e: Orbital period**

The period is:

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

Therefore:

\[
\frac{T_2}{T_1}
=
\frac{r_2}{r_1}
\frac{v_1}{v_2}
\]

Substituting the ratios:

\[
\frac{T_2}{T_1}
=
2\sqrt2
\]

\[
\frac{T_2}{T_1}\approx2.83
\]

The new satellite takes approximately 2.83 times as long to complete one revolution.

**Why the tempting approach fails**

A common mistake is to use \(h\), the altitude above Earth's surface, instead of \(r\), the distance from Earth's centre.

For example, the altitude increases from \(R_E\) to \(3R_E\), which is three times larger. But the actual orbital radius only doubles, from \(2R_E\) to \(4R_E\).

The inverse-square law applies to distance from the centre of the attracting mass, not distance above its surface.

This distinction becomes particularly important when comparing low and high satellite orbits.

### Question 5

<text> <em>Extension challenge: a satellite that receives a sudden increase in speed.</em> </text>

A satellite is initially travelling in a circular orbit of radius \(r\) around Earth.

Its original circular orbital speed is:

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

A rocket engine briefly fires in the direction of the satellite's motion, increasing its speed to:

\[
v'=1.20v_c
\]

The engine then switches off.

Assume:

- The satellite remains at the same distance \(r\) from Earth's centre during the brief impulse.
- The impulse is tangential to the original circular orbit.
- Earth's gravity is the only significant force after the engine switches off.
- Atmospheric drag and the gravitational effects of other bodies are negligible.

Answer the following:

a. Does the gravitational force acting on the satellite immediately after the impulse increase, decrease, or remain unchanged? Explain.

b. Does the satellite continue along its original circular orbit? Justify your answer mathematically.

c. Calculate the radius of curvature of the satellite's new trajectory immediately after the impulse, in terms of \(r\).

d. Will the satellite remain gravitationally bound to Earth, or escape? Explain your reasoning.

e. Explain why this situation exposes a mistake in treating centripetal force as an independent force that must increase whenever an object's speed increases.

[IMAGE: Earth with an initially circular satellite orbit. Show a satellite at the right-hand side of the orbit with its original tangential velocity arrow labelled vc and a second, longer tangential arrow labelled 1.20vc. Indicate that the speed increase occurs instantaneously at this position. Leave the subsequent trajectory and force arrows undrawn so students must reason them out.]

### Solution 5

The gravitational force remains unchanged, but the satellite cannot continue in its original circular orbit. Its immediate radius of curvature becomes \(1.44r\), and it remains gravitationally bound to Earth, following an elliptical orbit under the stated assumptions.

**Part a: Gravitational force**

Newton's law of gravitation is:

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

Immediately after the impulse, the satellite is still at the same distance \(r\) from Earth's centre.

Its mass is also unchanged.

Therefore, the gravitational force is unchanged.

Increasing the satellite's tangential speed does not directly increase Earth's gravitational attraction.

This is the first important distinction: gravitational force depends on mass and separation, not orbital speed.

**Part b: Does it remain in a circular orbit?**

For the original circular orbit:

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

After the speed increase, the inward acceleration required to maintain the original circular path would be:

\[
a_{\text{required}}
=
\frac{(1.20v_c)^2}{r}
\]

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

Therefore, the required acceleration would be 1.44 times its original value.

However, the actual gravitational acceleration is unchanged because the satellite remains at the same orbital radius.

It cannot supply the larger inward acceleration needed to maintain the original circular path.

The satellite therefore cannot remain in that circle.

This is where the tempting argument fails. A student might reason that a higher speed requires a larger centripetal force, so gravity must somehow become stronger.

But gravity does not change just because the satellite speeds up. Instead, the satellite changes its trajectory.

**Part c: New radius of curvature**

This part requires one additional idea from curved-path motion.

The radius of curvature, \(\rho\), describes how tightly a trajectory bends at a particular instant. For motion with acceleration perpendicular to velocity:

\[
a=\frac{v^2}{\rho}
\]

Immediately after the impulse, the gravitational acceleration is still:

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

The new speed is:

\[
v'=1.20v_c
\]

Therefore:

\[
\rho=\frac{v'^2}{a}
\]

Substituting:

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

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

Thus:

\[
\rho=1.44r
\]

The new trajectory is less tightly curved at that instant than the original circular path.

An important detail: the radius of curvature is not the same as the distance from Earth to the satellite once the path is no longer circular. Confusing those two quantities would produce an incorrect calculation.

**Part d: Does the satellite escape?**

The satellite remains gravitationally bound.

For an object in a circular orbit, the escape speed at the same radius is:

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

Since:

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

we can write:

\[
v_{\text{esc}}=\sqrt2v_c
\]

Numerically:

\[
v_{\text{esc}}\approx1.414v_c
\]

The new speed is:

\[
v'=1.20v_c
\]

Since:

\[
1.20v_c<1.414v_c
\]

the satellite does not have enough speed to escape Earth's gravitational field.

Instead, it follows an elliptical orbit. The point where the impulse occurs becomes the closest point of that new orbit, under the stated ideal conditions.

**Part e: The deeper lesson**

Centripetal force is not an independent force that increases automatically with speed.

It describes the net inward force required for a particular circular path.

Initially, gravity supplies exactly the inward force needed for a circular orbit.

After the impulse, gravity remains unchanged, but the satellite's velocity changes. The original circular path is no longer possible, so the trajectory changes.

This is the crucial distinction between a force determining an object's motion and an equation describing the force required for a particular motion.

The same reasoning becomes useful when studying elliptical orbits, orbital energy, and how rocket manoeuvres change a satellite's trajectory.