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# Gravity and Newton's Cannon

by Ron Kurtus (revised 7 March 2011)

* Newton's Cannon* (also called

*Newton's Cannonball*) was a "thought experiment" created by Isaac Newton in 1687 and stated in his

*Principia Mathematica*. In the book, he imagined shooting a cannonball parallel to the Earth's surface from the top of a very high mountain.

Depending on the velocity of the cannonball, it would strike the ground at some distance from the mountain top, go into orbit around the Earth or fly off into space. Of course, this assumes air resistance is negligible.

Questions you may have include:

- When would the cannonball hit the ground?
- How would the cannonball go into orbit?
- When would it fly off into space?

This lesson will answer those questions. Useful tool: Units Conversion

## Cannonball hits the ground

When a cannonball is fired parallel to the Earth's surface from a cannon on the top of a high mountain, the ball will usually travel for some distance until it hits the ground.

Assuming air resistance is negligible, the cannonball's displacement depends on its initial horizontal velocity and the time it takes the ball to fall to the ground from its given height.

Note: Displacement is how far the cannonball has traveled from the cannon along the horizontal axis. It is a vector in a specified direction. Distance is a scalar quantity that indicates how far the object traveled along its curved path.(

See Convention for Direction in Gravity Equations for more information.)

### When curvature not a factor

If the velocity of the cannonball is only sufficient to carry it several kilometers or miles, the curvature of Earth does not really come into play. The equation for its displacement is:

x = v_{h}√(2y/g)

where

**x**is the horizontal displacement in meters (m) or feet (ft)**v**is the initial horizontal velocity in m/s or ft/s_{h}**y**is the displacement the cannonball has fallen from its starting point in m or ft**g**is the acceleration due to gravity (9.8 m/s^{2}or 32 ft/s^{2})

(

See Effect of Gravity on Sideways Motion for more information.)

#### Would travel in parabolic path

You can see that if the ground is considered flat that the cannonball travels in a parabolic path by squaring both sides of the equation:

x^{2}= (2v_{h}^{2}/g)y

This results in the equation of a parabola, where the constant **k = 2v _{h}^{2}/g**:

x^{2}= ky

### When curvature is a factor

For greater distances, the curvature of the Earth becomes a factor and complicates the equation.

When the ground is considered flat, the force of gravity is perpendicular to the horizontal velocity of the cannonball. However, when the curvature of the Earth is taken into consideration, the direction of gravity changes with the distance traveled. It is assumed that the force of gravity is concentrated at the center of the Earth.

The cannonball travels in an elliptical path that is interrupted by the surface of the Earth. The lower axis of the ellipse is at the center of the Earth.

Cannonball follows elliptical path to hit the ground

As the initial velocity of the cannonball is increased, the ellipse becomes larger and the upper axis approaches the lower axis.

## Cannonball goes into orbit

At some initial velocity, the cannonball will not hit the ground but will go into orbit around the Earth.

### Initial elliptical orbit

As the initial velocity of the cannonball is increased, the size of the elliptical path increases to the point where the ball does not hit the ground and instead barely orbits the Earth barely orbits the Earth. The upper axis of the ellipse approaches the lower axis at the center of the Earth.

### Circular orbit

At a certain initial velocity of the cannonball, the upper axis and lower axis of the ellipse coincide at the center of the Earth. In other words, the elliptical orbit becomes a circular orbit. The initial velocity for a circular orbit is:

v_{c}= √(gR_{E})

where

**v**is the initial velocity of the cannonball required for a circular orbit in m/s_{c}**g**is the acceleration due to gravity = 9.8 m/s^{2}**R**is the radius of the Earth = 6.371*10_{E}^{6}m

Cannonball goes into a circular orbit

### Circular orbit example

What would be the required initial velocity of the cannonball to orbit the Earth?

#### Solution

Since the radius of the Earth is **R _{E}** = 6.371*10

^{6}m, the initial velocity is:

v9.8*6.371*10_{c}= √(^{6})

v62.436*10_{c}= √(^{6})

v7.902*10_{c}=^{3}m/s = 7902 m/s

This is much greater than the muzzle velocity of modern military cannons, which reach velocities of about 1,800 m/s. In fact the new electromagnetic railguns being developed by the U.S. Navy reach only 2400 m/s.

The result is that Newton's Cannon could work in theory, but there is no existing way to fire a cannonball or any projectile at the velocity required to orbit the Earth.

### Elliptical orbit

As the initial velocity of the cannonball increases, the orbit becomes elliptical again. The upper axis of the ellipse is now at the Earth's center. The lower axis moves further away as the initial velocity is increased.

Cannonball goes into a large elliptical orbit around the Earth

The velocity of the ball is for these elliptical paths greater than that for a circular path but less than the escape velocity:

√(gR_{E}) < v_{c}< √(2gR_{E})

where

**√(gR**means that_{E}) < v_{c}**v**is greater than the velocity for a circular orbit_{c}**v**means that_{c}< √(2gR_{E})**v**is less than the escape velocity from gravity_{c}

(

See Escape Velocity from Gravity for more information.)

## Cannonball goes into space

The cannonball will fly off into space, if its initial velocity is:

v_{e}≥ √(2gR_{E})

where

**v**is the escape velocity from Earth's gravity_{e}**≥**means greater than or equal to...

If **v _{e} = √(2gR_{E})**, the path of the cannonball is parabolic.

If **v _{e} > √(2gR_{E})**, the path of the cannonball is hyperbolic.

(

See Effect of Velocity on Orbital Motion for more information.)

## Summary

*Newton's Cannon* was a "thought experiment" created by Isaac Newton where he imagined shooting a cannonball parallel to the Earth's surface from the top of a very high mountain.

Depending on the velocity of the cannonball, it would strike the ground at some displacement from the mountain top, go into orbit around the Earth or fly off into space. The paths of the cannonball would be elliptical until the velocity reaches the escape velocity, where the path would become parabolic or hyperbolic.

However, the required velocities for orbiting the Earth or going into space are really larger than possible for even modern cannons.

Use your imagination to see new things

## Resources and references

### Websites

**Newton's Cannonball and the Speed of Orbiting Objects** - Bucknell University Physics

**Newton's cannonball** - Wikipedia

**Interactive Demonstration of Newton's Cannon**

### Books

**Top-rated books on Simple Gravity Science**

**Top-rated books on Advanced Gravity Physics**

## Questions and comments

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## Gravity and Newton's Cannon