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Effect of Gravity on Sideways Motion

by Ron Kurtus (revised 21 January 2011)

When an object is moving sideways, horizontal or parallel to the Earth's surface at a constant velocity, the effect of gravity on the object is independent of the object's lateral movement.

In other words, an object moving sideways will fall at the same rate as one that is simply dropped. You can derive the equation for the displacement of the object before it hit the ground from the gravity equations for falling objects. An exception is if the object moves so far that the curvature of the Earth comes into play during its fall to the ground.

Questions you may have include:

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



Gravity does not affect sideways motion

The force of gravity acts on objects in a direction that is perpendicular to level ground.

Note: Since the Earth is a sphere with a circumference of approximately 40,000 km or 25,000 mi, the ground can be considered level or flat (not counting hills and valleys) for short displacements of several kilometers or several miles.

This means that, if an object is moving parallel to the ground, the force of gravity is only pulling on the object in a downward direction. The force is not affected by sideways motion and simply pulls the object down at the same rate as if it was stationary.

This rule is explained in detail in Horizontal Motion Unaffected by Gravity.

Pull of gravity downward is the same for moving and stationary objects

Pull of gravity downward is the same for moving and stationary objects

Objects hit ground at same time

If you would project or throw an object exactly parallel to the Earth's surface, the sideways motion of the object would have no effect on how gravity acts on it. In other words, the object would drop at the same rate as an object dropped from the identical height. The time it would take either object to hit the ground would be the same.

Ball thrown sideways falls at the same rate as dropped ball

Ball thrown sideways falls at the same rate as dropped ball

Simple experiment

You can try a simple experiment to verify this phenomenon. Place a coin on the edge of a table or desk and hold another coin at the same height. With one hand flick the coin on the table across the room. At the same time, drop the other coin. You will hear that they hit the floor at just about the same time.

Derivation of equation for displacement

You can find the displacement an object projected sideways from the following derivation. First, consider the standard displacement-velocity equation:

x = vst

where

The equation for the time a falling object takes is:

t = √(2y/g)

where

(See Gravity Time Equations for Falling Objects for information on the equation.)

Thus, the displacement the object travels, as a function of the initial velocity and the height is:

x = vs√(2y/g)

Travels in parabolic curve

You can see that object travels in a parabolic path by squaring both sides of the equation, with x as the vertical axis and y as the horizontal axis:

x2 = (2vs2/g)y

This results in the equation of a parabola, where the constant k = 2vh2/g:

x2 = ky

Shooting and dropping a bullet

If you would shoot a bullet from a gun exactly parallel to the Earth's surface, the motion of the bullet would have no effect on how gravity acts on the bullet. In other words, the bullet would drop at the same rate as a stationary object.

Dropped bullet and shot bullet hit the ground at the same time

Dropped bullet and shot bullet hit the ground at the same time

Many people don't believe that if you held a rifle or handgun parallel to the ground and at the same time you shot the bullet, you dropped another bullet from the same height, both bullets would both hit the ground at the same time. However, it is a fact.

Exception

An exception to this phenomenon would be if the bullet or object was able to travel so many miles or kilometers that the curvature of the Earth came into play. In such a situation, the bullet would take slightly longer to hit the ground, because the displacement to the ground was greater due to the Earth's curvature.

Example

If you shot a bullet at 900 m/s from a rifle that was 1.5 m above the ground, how far would the bullet fly until it hit the ground? Discount air resistance and assume the rifle is parallel to the ground.

Solution

x = vs√(2y/g)

x = (900 m/s)[2*(1.5 m)/(9.8 m/s2)]

x = (900 m/s)(0.306 s2)

x = (900 m/s)(0.553 s)

x = 498 m or 1634 ft

Summary

An object moving sideways or parallel to the Earth's surface will fall at the same rate as one that is simply dropped. The equation for the displacement of the object before it hits the ground can be derived from the gravity equations for falling objects. An exception is if the object moves so fast or far that the curvature of the Earth comes into play during its fall to the ground.


Be curious about the world around you


Resources and references

Ron Kurtus' Credentials

Websites

Independence of Perpendicular Components of Motion - PhysicsClassroon.com

Gravity Resources

Books

Top-rated books on Simple Gravity Science

Top-rated books on Advanced Gravity Physics


Questions and comments

Do you have any questions, comments, or opinions on this subject? If so, send an email with your feedback. I will try to get back to you as soon as possible.


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