1. I wanted to know how far an object would go if I threw it in a perfect vacuum with only the force of gravity acting upon it, and seeing as I could not find an equation on the internet, I came up with this:

dA = Fa/F

where

dA = distance attained
Fa = force applied
F = Value of specified object in Newton's Law of Universal Gravitation

I made an example and did a test, which seemed relatively successfully in accordance to the answer to the equation, in account of no vacuum and other more hidden factors.

dA = Fa/F

(Example for earth)
An object of 1kg thrown with a force of 100N 1m from the ground on a perfect body the with a diameter roughly proportional to that of the earth itself.

dA = 100/ 0.000000000005339072 X 1kg x 59736000000000000000000000kG/1^2

If calculation is correct, the object in question should attain a total distance of

1.118846122m ( My experiment involved a football with 100N of force applied, thrown from 1m high and attaining roughly 1m distance.  2.

3. You don't think the angle (relative to the ground) at which it's thrown is relevant?  4. Originally Posted by Devon Keogh I wanted to know how far an object would go if I threw it in a perfect vacuum with only the force of gravity acting upon it, and seeing as I could not find an equation on the internet,
Range of a projectile - Wikipedia, the free encyclopedia

• g: the gravitational acceleration—usually taken to be 9.81 m/s2 (32 f/s2) near the Earth's surface
• θ: the angle at which the projectile is launched
• v: the velocity at which the projectile is launched
• y0: the initial height of the projectile
• d: the total horizontal distance travelled by the projectile
When neglecting air resistance, the range of a projectile will be If (y0) is taken to be zero, meaning the object is being launched on flat ground, the range of the projectile will then simplify to   5. Originally Posted by Dywyddyr You don't think the angle (relative to the ground) at which it's thrown is relevant?
No, it is always a straight throw.  6. Originally Posted by Devon Keogh (Example for earth)
An object of 1kg thrown with a force of 100N 1m from the ground on a perfect body the with a diameter roughly proportional to that of the earth itself.

the statement "thrown with a force of 100N" has no meaning by itself. You have to supply the distance over which the force was applied. For example, if while throwing the body, you applied 100N to the object for 1/2 meter from start of the throw until release, it will travel 1.414 times further than if you had shortened your throw and applied the same force for 1/4 meter.

dA = 100/ 0.000000000005339072 X 1kg x 59736000000000000000000000kG/1^2

If calculation is correct, the object in question should attain a total distance of

1.118846122m ( My experiment involved a football with 100N of force applied, thrown from 1m high and attaining roughly 1m distance.
One example does not make a case for the equation. You would have to make several throws, at different heights and with different forces and check them all again the equation.

And even then you could only say that the formula was sufficiently accurate over the range tested.

Here's what I mean: At low enough throwing speeds and low enough heights you can get accurate estimates of distance traveled by assuming the the ground is perfectly flat, gravity does not change with height, and the force of gravity is parallel to itself at all points of the trajectory.

In this case all you have to do is determine how much time it would take for the body to fall to the ground is dropped from its height and multiple this by the throwing speed to get the distance traveled. Assuming that the throw is made parallel to the ground.

However at high enough heights and speeds, you have to take into account that gravity weakens with altitude, the curvature of the Earth's surface, and that gravity acts towards the center of the Earth.

In these cases, the solution is more involved.

First you find the total orbital energy of the thrown body (kinetic + potential energy)
You use this to find the semi-major axis of the orbital path
Now you can find the eccentricity of the path
With the eccentricity, the radius of the Earth and height above the surface the throw is made, you can find the angle between the throwing point and landing point as measured from the center of the Earth.
This and radius of the Earth will give you the distance the body traveled before hitting.

And even the above solution ignores corrections that have to be made due to the rotation of the Earth.  7. Surly once the object is in the vacuum the vacuum is no longer a vacuum?  8. Originally Posted by andythomasthekey Surly once the object is in the vacuum the vacuum is no longer a vacuum?
Then, the innumerable man-made objects now in orbit about the Earth, are NOT in a vacuum? jocular  9. Does that mean Earth is also in the vacuum, or has it made a little space for its self within the vacuum?  10. Originally Posted by andythomasthekey Does that mean Earth is also in the vacuum, or has it made a little space for its self within the vacuum?
Stop splitting hairs. The point is, there is no air friction on the projectile. Arguing about the definition of vacuum is just vacuous.  11. Sorry if I seem to be disrupting the thread, I just can't picture what is being said as there is no such thing as no friction on a projectile, as far as I am aware.  12. Originally Posted by andythomasthekey Sorry if I seem to be disrupting the thread, I just can't picture what is being said as there is no such thing as no friction on a projectile, as far as I am aware.
If there isn't any air, where does the friction come from? In order to get an accurate result, it doesn't have to be identically zero, just small enough for your required accuracy. In many cases, air friction may be ignored even in normal atmosphere. It all depends on the conditions of your experiment.  Bookmarks
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