Why Orbiting Spacecraft Keep Falling Without Hitting Earth

An orbiting spacecraft about 400 km above Earth is still being pulled towards Earth’s centre by gravity. In fact, its gravitational acceleration is approximately \(8.7\ \mathrm{m\,s^{-2}}\), which is nearly 89% of the gravitational acceleration at Earth’s surface.

So why doesn’t it fall straight down and crash?

Because it is travelling sideways at approximately \(7.7\ \mathrm{km\,s^{-1}}\). Gravity continually bends its path towards Earth, while its sideways motion carries it forwards. The spacecraft is in continuous free fall, but its path curves around Earth rather than intersecting the surface.

Here’s the important idea: an orbiting spacecraft doesn’t avoid falling. It keeps falling, but it keeps missing Earth.

Predict what would happen in two different situations. If gravity suddenly disappeared, would the spacecraft continue following its circular orbit? No. It would travel in a straight line along the tangent to its orbit. If its sideways velocity disappeared instead, would it remain in orbit? No. It would accelerate directly towards Earth.

An orbit requires both gravitational acceleration and the right sideways velocity.

01What does it mean to keep falling?

Start by separating two ideas that are easy to confuse: velocity and acceleration.

Velocity describes how quickly an object moves and in which direction. Acceleration describes how quickly its velocity changes.

An object can travel at a constant speed and still accelerate if its direction changes.

Think about a car travelling around a roundabout at a steady 40 km/h. Its speed stays constant, but its direction changes continuously. The tyres exert an inward force that makes the car follow the curved path.

A spacecraft behaves similarly, except that gravity provides the inward force instead of friction between tyres and a road.

The analogy has a limit: a car needs contact with the road to produce the inward force. An ideal spacecraft in orbit needs no such support. Gravity alone provides the required acceleration.

The two directions that matter

At any instant during a circular orbit, the spacecraft has:

• A velocity directed along the tangent to its circular path.
• A gravitational acceleration directed towards Earth’s centre.

These directions are perpendicular to each other.

The spacecraft’s velocity carries it forwards, while gravity continuously changes the direction of that velocity.

This inward acceleration is called centripetal acceleration. The word centripetal means centre-seeking.

It is not an additional force that somehow appears in space. It describes the acceleration required for circular motion. In an ideal Earth orbit, gravity supplies the entire centripetal force.

In a circular orbit, the gravitational acceleration is perpendicular to the spacecraft’s velocity. This means gravity changes the direction of motion without changing the speed.

That explains something that might initially seem contradictory: a spacecraft can have constant speed and non-zero acceleration at the same time.

Its altitude remains constant because its distance from Earth’s centre remains constant. It is not hovering. It is moving continuously around Earth under the influence of gravity.

02Newton’s Moon test: does gravity act beyond Earth’s surface?

The idea that an object could continually fall without reaching the ground was part of a much bigger problem in the development of physics.

Why does an apple fall towards Earth, while the Moon continues moving around Earth instead of crashing into it?

Isaac Newton investigated whether the same gravitational attraction responsible for objects falling near Earth’s surface could also explain the Moon’s curved path.

His mathematical work was developed during the 1680s and published in Philosophiæ Naturalis Principia Mathematica in 1687. The earlier De motu work and Newton’s discussions with Edmund Halley formed part of this development. The Newton Project documents this historical progression.

Importantly, this was not a laboratory experiment in which Newton launched an object into space. It was a mathematical comparison using terrestrial measurements and astronomical observations.

What information was available?

The reasoning required two sets of information.

Near Earth’s surface:

A falling object’s acceleration due to gravity was approximately:

\[ g_E=9.8\ \mathrm{m\,s^{-2}} \]

At the Moon:

Astronomical observations provided an estimate of the Moon’s orbital distance and its time to complete one revolution.

The Moon’s orbital radius is approximately 60 times Earth’s radius, and its orbital period is approximately 27.3 days.

What should Newton’s model predict?

Newton’s gravitational model predicts that gravitational acceleration decreases with the square of the distance from Earth’s centre.

If the Moon is 60 times farther from Earth’s centre than an object at the surface, its gravitational acceleration should be:

\[ a_M=\frac{g_E}{60^2} \]

Therefore,

\[ a_M=\frac{9.8}{3600} \]

\[ a_M\approx 0.00272\ \mathrm{m\,s^{-2}} \]

The prediction is that the Moon experiences a small but definite acceleration towards Earth.

What did the observations suggest?

The Moon’s orbital radius and period could be used to calculate the centripetal acceleration required for its approximately circular motion.

The resulting value is also close to:

\[ a_M\approx 0.0027\ \mathrm{m\,s^{-2}} \]

This agreement supported the idea that the same gravitational law explains both falling objects near Earth and the Moon’s orbital motion.

The Moon does not escape gravity. Gravity continually bends its trajectory towards Earth.

The calculation is an idealised comparison. The Moon’s actual orbit is slightly elliptical, and the Sun also influences its motion. A more complete model must account for these effects.

The important historical conclusion is that terrestrial and celestial motion did not need separate explanations. The same gravitational interaction could account for both.

A modern discussion of this comparison is available in OpenStax’s treatment of Newton’s universal law of gravitation.

03The circular-orbit condition

We can now apply the same reasoning to a spacecraft orbiting Earth.

There are two forces and motion relationships to connect.

First, Newton’s law of universal gravitation gives the magnitude of the gravitational force:

\[ F_g=\frac{GM_E m}{r^2} \]

Second, Newton’s second law tells us that the net force required for circular motion is:

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

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

Therefore:

\[ \frac{GM_E m}{r^2}=\frac{mv^2}{r} \]

Notice that the spacecraft’s mass, \(m\), appears on both sides.

It cancels, leaving:

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

This gives two useful relationships.

The gravitational acceleration is:

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

The speed required for a circular orbit is:

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

Here:

• \(G\) is the universal gravitational constant, \(6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\).
• \(M_E\) is Earth’s mass, approximately \(5.97\times10^{24}\ \mathrm{kg}\).
• \(m\) is the spacecraft’s mass, measured in kilograms.
• \(r\) is the orbital radius, measured from Earth’s centre, in metres.
• \(v_c\) is the circular orbital speed, measured in metres per second.
• \(a_c\) is the magnitude of the centripetal acceleration, measured in metres per second squared.

Both acceleration and gravitational force point towards Earth’s centre.

Don’t confuse altitude with orbital radius

This is an important calculation detail.

The altitude \(h\) is the spacecraft’s height above Earth’s surface. The orbital radius \(r\) is its distance from Earth’s centre.

They are related by:

\[ r=R_E+h \]

where \(R_E\) is Earth’s radius.

You must use \(r\), not \(h\), in the orbital equations.

The equations above assume:

• A circular orbit around Earth.
• Earth can be treated as a spherical body with its mass concentrated at its centre.
• The spacecraft’s mass is negligible compared with Earth’s mass.
• Atmospheric drag, thrust, and gravitational disturbances from other bodies are ignored.

Real spacecraft can experience small orbital changes due to these neglected effects.

04Why astronauts float inside an orbiting spacecraft

If gravity is still acting so strongly, why do astronauts and loose objects float inside an orbiting spacecraft?

The answer is that everything inside the spacecraft is falling together.

Imagine an elevator whose cable suddenly breaks. If you and a small object inside it were released at the same time, you would both accelerate downwards together.

The floor would no longer push upwards against your feet. The object would appear to float beside you.

An orbiting spacecraft creates a similar situation, except that the entire cabin is falling around Earth rather than towards the ground.

Consider three objects:

1. The spacecraft.
2. An astronaut inside it.
3. A loose tool released by the astronaut.

If we ignore small differences in gravitational acceleration across the cabin, all three experience approximately the same gravitational acceleration towards Earth’s centre.

The astronaut and tool therefore remain approximately stationary relative to the cabin.

Does the astronaut actually have zero weight?

Not in the sense that gravity has stopped acting.

At 400 km altitude, a 60 kg astronaut experiences a gravitational force of approximately:

\[ F_g=mg \]

\[ F_g=(60)(8.69) \]

\[ F_g\approx521\ \mathrm{N} \]

The gravitational force is still substantial.

However, the astronaut does not feel the usual supporting force from the floor. A spring balance beneath their feet would read approximately zero because the astronaut and the balance are accelerating together.

This is called apparent weightlessness.

The distinction is important:

• Gravitational force: still acts towards Earth.
• Apparent weight: approximately zero because there is no supporting contact force.

NASA explains this distinction through the free-fall motion of spacecraft, astronauts, and objects inside them in its Basics of Space Flight resource.

05The contrasting case: being high above Earth isn’t enough

Imagine placing a spacecraft 400 km above Earth’s surface.

Does it automatically enter orbit?

No.

Its altitude alone does not determine whether it will orbit Earth. Its velocity matters just as much.

Consider two spacecraft at the same altitude.

Spacecraft A has the correct tangential speed for a circular orbit. Gravity continuously changes its direction, keeping its distance from Earth’s centre constant.

Spacecraft B has no tangential velocity. Gravity accelerates it directly towards Earth.

Both spacecraft experience approximately the same gravitational acceleration at that altitude. Their different paths result from their different initial velocities.

This exposes a common misconception: sideways velocity does not cancel gravity.

Gravity is not balanced by a horizontal force. It remains the net force acting on the spacecraft and continuously accelerates it inwards.

The sideways velocity determines the path that the spacecraft follows under that acceleration.

A useful decision process for circular-orbit questions

When solving an HSC question about an orbiting spacecraft, follow this sequence:

1. Find the orbital radius.

If the question gives altitude:

\[ r=R_E+h \]

2. Calculate gravitational acceleration.

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

Remember that its direction is towards Earth’s centre.

3. Determine the required circular speed.

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

4. Check the velocity direction.

For a circular orbit, the velocity must be tangential to the orbit.

The correct speed with the wrong direction does not produce the circular motion described by these equations.

5. Explain the physical meaning.

A strong written response should connect the inward gravitational acceleration with the spacecraft’s tangential velocity. State that gravity provides the centripetal force and that the spacecraft remains in continuous free fall.

06Questions and solutions

Question 1

An Earth-observation satellite follows a circular orbit 550 km above Earth’s surface.

Use:

• \(R_E=6.37\times10^6\ \mathrm{m}\)
• \(M_E=5.97\times10^{24}\ \mathrm{kg}\)
• \(G=6.67\times10^{-11}\ \mathrm{N\,m^2\,kg^{-2}}\)

(a) Calculate its orbital radius.

(b) Calculate its gravitational acceleration, including its direction.

(c) Explain why this acceleration does not mean the satellite is losing altitude.

Solution 1

The satellite’s orbital radius is \(6.92\times10^6\ \mathrm{m}\), and its gravitational acceleration is \(8.32\ \mathrm{m\,s^{-2}}\) towards Earth’s centre. Its altitude remains constant because it is following a circular orbit.

Step 1: Calculate the orbital radius.

\[ r=R_E+h \]

\[ r=6.37\times10^6+5.50\times10^5 \]

\[ r=6.92\times10^6\ \mathrm{m} \]

Step 2: Calculate gravitational acceleration.

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

\[ a_g= \frac{(6.67\times10^{-11})(5.97\times10^{24})} {(6.92\times10^6)^2} \]

\[ \boxed{a_g=8.32\ \mathrm{m\,s^{-2}}} \]

The acceleration points towards Earth’s centre.

Step 3: Explain why the altitude stays constant.

The satellite’s velocity is tangential to its circular path, while its acceleration is directed inwards.

Gravity continually changes the direction of the satellite’s velocity, producing circular motion.

The distance from Earth’s centre therefore remains constant.

Question 2

A 900 kg satellite and a 60 kg instrument capsule are travelling in separate circular orbits at the same altitude of 400 km above Earth.

(a) Calculate the gravitational force acting on each object.

(b) Compare their centripetal accelerations.

(c) A student argues that the heavier satellite must have a greater centripetal acceleration because Earth’s gravitational force on it is greater. Explain the error in this reasoning.

Solution 2

The 900 kg satellite experiences a greater gravitational force, but both objects have the same centripetal acceleration of \(8.69\ \mathrm{m\,s^{-2}}\).

Step 1: Calculate gravitational acceleration at 400 km.

From the earlier calculation:

\[ a_g=8.69\ \mathrm{m\,s^{-2}} \]

Step 2: Calculate the gravitational force on the 900 kg satellite.

Using Newton’s second law:

\[ F_g=ma_g \]

\[ F_g=(900)(8.69) \]

\[ \boxed{F_g=7.82\times10^3\ \mathrm{N}} \]

Step 3: Calculate the gravitational force on the 60 kg capsule.

\[ F_g=(60)(8.69) \]

\[ \boxed{F_g=5.21\times10^2\ \mathrm{N}} \]

Step 4: Compare their accelerations.

Newton’s second law gives:

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

For the heavier satellite, the gravitational force is larger because its mass is larger.

However, its inertia is also proportionally larger.

The mass cancels when gravitational force is divided by mass:

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

Both objects therefore experience the same acceleration at the same orbital radius.

The student’s mistake is confusing a larger force with a larger acceleration without accounting for the object’s mass.

Question 3

An astronaut is inside a spacecraft travelling in a circular orbit 400 km above Earth’s surface.

The astronaut releases a 0.20 kg calibration mass. It is released with zero velocity relative to the spacecraft.

Assume the gravitational field is uniform across the small spacecraft.

(a) What is the calibration mass’s acceleration in an Earth-centred reference frame?

(b) Will the calibration mass fall towards the cabin floor?

(c) If the astronaut stands on a spring balance attached to the cabin floor, what will the balance read?

Explain your answers.

Solution 3

The calibration mass accelerates towards Earth’s centre at \(8.69\ \mathrm{m\,s^{-2}}\), does not fall towards the cabin floor in the stated approximation, and the spring balance reads approximately zero.

Step 1: Determine the gravitational acceleration.

The mass and spacecraft are at the same orbital altitude.

Therefore, the gravitational acceleration is:

\[ a_g=8.69\ \mathrm{m\,s^{-2}} \]

directed towards Earth’s centre.

Step 2: Consider the motion relative to the spacecraft.

The calibration mass initially has the same velocity as the spacecraft.

Once released, both the spacecraft and calibration mass continue accelerating under gravity.

Because the question specifies a uniform gravitational field across the cabin, they experience the same acceleration.

There is therefore no relative acceleration between them in this approximation.

The calibration mass appears to float beside the astronaut rather than falling towards the floor.

Step 3: Determine the spring balance reading.

A spring balance measures the contact force supporting the astronaut.

In free fall, the astronaut and balance accelerate together. The balance does not need to exert an upward supporting force.

Therefore:

\[ \boxed{N\approx0\ \mathrm{N}} \]

The gravitational force has not disappeared.

The astronaut, spacecraft, and calibration mass are simply falling together.

Question 4

Two identical capsules are placed at the same position, 400 km above Earth’s surface.

Capsule A is travelling tangentially at \(7.67\ \mathrm{km\,s^{-1}}\).

Capsule B has zero velocity relative to an Earth-centred inertial reference frame.

Ignore Earth’s rotation, atmospheric drag, and gravitational influences from other bodies.

(a) Calculate the gravitational acceleration experienced by each capsule at the instant described.

(b) Predict the subsequent motion of each capsule.

(c) Explain why the two capsules have the same initial gravitational acceleration but follow different paths.

Solution 4

Both capsules initially experience a gravitational acceleration of \(8.69\ \mathrm{m\,s^{-2}}\) towards Earth’s centre. Capsule A follows a circular orbit, while Capsule B falls radially towards Earth.

Step 1: Calculate the gravitational acceleration.

Both capsules have the same orbital radius:

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

Therefore:

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

\[ \boxed{a_g=8.69\ \mathrm{m\,s^{-2}}} \]

This acceleration is directed towards Earth’s centre for both capsules.

Step 2: Analyse Capsule A.

Its tangential speed is:

\[ v=7.67\times10^3\ \mathrm{m\,s^{-1}} \]

This is the circular orbital speed at the given radius.

Gravity supplies the centripetal acceleration required for circular motion.

Therefore, Capsule A continues around Earth at a constant orbital radius.

Step 3: Analyse Capsule B.

Capsule B has no initial tangential velocity.

Its gravitational acceleration is directed towards Earth’s centre, so it begins falling radially towards Earth.

It does not have the sideways motion required for a circular orbit.

Step 4: Explain the difference.

Gravitational acceleration depends on the central mass and the object’s distance from its centre.

Both capsules have the same initial position, so they experience the same gravitational acceleration.

However, acceleration alone does not determine the complete motion. Initial velocity also matters.

This is why one capsule follows a circular orbit while the other falls towards Earth.

Question 5

An advanced spacecraft follows a circular orbit around Earth with an orbital radius of:

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

A 10.0 m rigid boom extends directly away from Earth’s centre. A small sensor pod is attached to the outer end.

The pod is released without any additional push.

At the instant of release:

• The pod has the same tangential velocity vector as the spacecraft’s centre.
• The pod has zero radial velocity.
• The pod is 10.0 m farther from Earth’s centre than the spacecraft’s centre.

Assume Earth is a spherical central mass, and ignore atmospheric drag, the boom’s mass, and gravitational influences from other bodies.

(a) Calculate the circular orbital speed of the spacecraft’s centre.

(b) Calculate the circular orbital speed required at the pod’s position.

(c) Determine whether the pod initially moves towards Earth or towards a greater orbital radius.

(d) Explain why your conclusion does not mean that gravity acts outwards on the pod.

Solution 5

The sensor pod initially moves towards a greater orbital radius. Although it has the same tangential speed as the spacecraft, that speed is slightly greater than the circular speed required at its larger radius.

This is a subtle example of why sharing the same velocity does not automatically mean two objects follow the same orbit.

Step 1: Calculate the spacecraft’s circular orbital speed.

Use:

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

Substitute:

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

Therefore:

\[ \boxed{v_c=7669.30\ \mathrm{m\,s^{-1}}} \]

The pod initially has this same tangential speed.

Step 2: Calculate the pod’s orbital radius.

The pod is 10.0 m farther from Earth’s centre.

Therefore:

\[ r'=r+10.0 \]

\[ r'=6.77001\times10^6\ \mathrm{m} \]

Step 3: Calculate the circular speed required at the pod’s position.

Use:

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

Substitute:

\[ v_c'= \sqrt{ \frac{3.98199\times10^{14}} {6.77001\times10^6} } \]

Therefore:

\[ \boxed{v_c'=7669.2947\ \mathrm{m\,s^{-1}}} \]

Step 4: Compare the actual speed with the required speed.

The pod’s actual tangential speed is:

\[ v_{\text{pod}}=7669.3004\ \mathrm{m\,s^{-1}} \]

Its required circular speed is:

\[ v_c'=7669.2947\ \mathrm{m\,s^{-1}} \]

The difference is:

\[ \Delta v=7669.3004-7669.2947 \]

\[ \boxed{\Delta v\approx0.0057\ \mathrm{m\,s^{-1}}} \]

The pod is travelling approximately 5.7 mm/s faster than the circular speed required at its own radius.

Step 5: Determine the direction of its initial motion.

For a circular orbit, gravity must provide exactly the inward acceleration required to maintain a constant radius.

At the pod’s position, its actual tangential speed is slightly greater than the required circular speed.

Consequently, the gravitational acceleration is insufficient to bend its path as tightly as a circular orbit at that radius would require.

The pod begins moving outwards, increasing its distance from Earth’s centre.

In the ideal two-body model, it follows a slightly elliptical orbit, initially moving away from Earth from its release point.

Why is this result surprising?

The tempting answer is that the pod must follow the spacecraft’s original circular orbit because it has exactly the same velocity at release.

But an object’s orbital motion depends on both its position and its velocity.

The pod starts at a different radius, where the circular orbital speed is slightly lower. Its unchanged tangential speed is therefore too high for a circular orbit at that position.

This also reveals a limitation of the uniform-field approximation used in Question 3. Across a small spacecraft cabin, differences in gravitational acceleration are usually negligible. Across larger separations, those differences can produce relative motion.

The key distinction is that gravity still points towards Earth. The pod’s increasing orbital radius results from its motion, not from an outward gravitational force.

07Where this leads next

An orbiting spacecraft is not suspended above Earth, and it is not outside Earth’s gravitational influence.

It is in continuous free fall. Its tangential velocity allows it to travel around Earth while gravity continually changes the direction of its motion.

Once you understand this, the next useful question is what happens when the spacecraft’s velocity changes. It can no longer maintain its original circular orbit, and its subsequent path must be analysed differently.

For more practice with the inward acceleration calculations, see How to Calculate a Satellite’s Centripetal Acceleration.

To investigate how a change in velocity alters an orbit, continue with How a Sudden Speed Change Affects a Satellite’s Orbit.