Geometry to work out how fast a bullet would have to travel to

[Hazzattack]

Hi guys, i was wondering how you calculate how fast a bullet would have to travel to 'overcome' gravity, i.e if it didn't hit anything (and discluding air resistance for now).
Mapping the Earth as a circle and knowing things such as the radius of the Earth and gravity acts on the bullet at 9.81 m/s^2 - how would i draw the triangles inside the circle to calculate the speed the bullet needs to travel to continue moving around the Earth (i.e the surface drops off at the same rate as the acceleration of gravity).

I wanted to solve the problem using geometry - which I'm aware would be a crude approximation. I couldn't find anything helpful via searching the internet, just continuous about which hits the ground first... blablabla.

Thanks a lot for any help,

 

[Redbelly98]

Are you familiar with the equation for centripetal force, i.e. the force required to keep an object moving in a circular path at constant speed? That force is $\frac{mv^2}{r}$, where m is the object's mass, v is it's speed, and r is the radius of the circular path.

Set that force equal to the gravitational force acting on the object -- since it is gravity that provides the necessary force -- and then you can figure out what v is.

p.s. I am not sure how drawing triangles would help out here.

 

[K^2]

There is a way of deriving mv²/r geometrically, which does involve triangles. It involves taking a limit as the final step, but it's a pretty simple proof. Could that be part of what OP is referring to?

 

[Hazzattack]

I am familiar with it, and admittedly, your way of solving the problem is far simpler. I found it in feynmanns lectures, chapter 7 I think, just after he talks about keplers laws. He solves this problem via geometry, but I found the calculation a little confusing (not due to its complexity). I'll take a picture of it later and upload it. Thanks for the responses.

 

[PJC01]

I would approach this problem by first understanding the basic principles of projectile motion and gravity. The equation for projectile motion is:

d = v0t + 1/2at^2

Where d is the distance traveled, v0 is the initial velocity, a is the acceleration due to gravity (9.81 m/s^2 on Earth), and t is the time.

In this case, we are interested in finding the velocity needed to overcome gravity and continue moving at a constant distance from the surface of the Earth. This can be calculated by setting the distance traveled (d) to be equal to the radius of the Earth (r).

Therefore, the equation becomes:

r = v0t + 1/2at^2

We also know that the time (t) it takes for the bullet to travel around the Earth is equal to the circumference of the Earth (2πr) divided by the velocity (v).

So, we can rewrite the equation as:

r = v0(2πr/v) + 1/2a(2πr/v)^2

Solving for v0, we get:

v0 = √(g * r)

Where g is the acceleration due to gravity and r is the radius of the Earth.

So, to find the velocity needed to overcome gravity and continue moving at a constant distance from the surface of the Earth, we simply need to take the square root of the product of the acceleration due to gravity and the radius of the Earth.

This calculation will give us the minimum velocity needed for the bullet to continue moving around the Earth without hitting the ground. However, this does not take into account air resistance, which would affect the actual velocity needed.

In conclusion, by using the principles of projectile motion and understanding the geometry of the Earth, we can calculate the velocity needed for a bullet to overcome gravity and continue moving around the Earth's surface. However, this is a simplified calculation and does not account for real-world factors such as air resistance. Further research and experimentation would be needed for a more accurate calculation.