Velocity to land 1/4, 1/2, and around Earth

[supernova88]

Hi, I'm actually a teacher presently undergoing gravitation and circular motion with my students, and ran into a problem I forgot how to think through. I'm trying to explain Newton's cannonball thought process (shoot a cannonball fast enough off a mountain and it will circle the earth), and determine the velocity v from a particular height. I completely understand the rationale behind v=sqrt(G x mass of Earth / radius), but my students aren't ready for that *yet*. Instead I wanted to simplify things and pretend Earth is a long flat surface (I know it isn't) where its circumference is its distance which is d = 40,000,000 m. We would then find the velocity to land 1/4, 1/2, and all the way "around" earth.

Assuming a mountain is 10 km or 10,000 m tall, freefall should take roughly 44.7 seconds if g ~10 m/s^2, and rounding up we get 50 s (yes I know I'm taking a lot of shortcuts but at this level it's necessary). Anyway, the obvious answer is that for a cannonball to fall 10 km and land 40,000,000 m away it would need to travel v = 40,000,000/50 = 800,000 m/s, which equals nearly 2 million mi/h. This is way bigger than the known orbital velocity of things like the space station which are around 17,000 mi/h.

Even traveling half way around Earth (20,000,000 m) the velocity is basically 1 million mi/h.

I did have some luck when instead I looked at a quarter of the distance around Earth and thought of it like a triangle, where the vertical leg (say, facing north) is Earth's radius plus the mountain, and the horizontal leg (east) is Earth's radius. The height the cannonball falls therefore 6,370 km + 10 km, and freefall takes 1128.7 s. Landing 6,370,000 m away to the east, it should go - approximately - 5,643.7 m/s = 20,317.2 km/h = 12,700 mi/h. This isn't quite 17,000 mi/h, but a step in the right direction.

Anyway, that's a long winded way of asking how I can simply demonstrate a cannonball's velocity to land 1/4, 1/2, and all the way around the earth.

Thanks so much in advance.

 

[supernova88]

So I just realized the error in my thinking ... orbits don't take 50 seconds!

I'm still curious for some simple ways to find the velocity it takes to land 1/4, 1/2, and 1 whole circle around the Earth. Thanks.

 

[BvU]

Your $g$ is permanently pointing in the same direction.
real $g$ for an orbit [edit] a circular orbit is permanently pointing perpendicular to the trajectory.
Makes a difference !

THere should be a better way to bring in Newton's cannonball, but I can't oversee what resources there are available.

I wouldn't dream of misleading my students in the way you propose -- too much risk of confusing average students (and losing your credibility with the bright ones !).

But then again, I don't teach .

 

[supernova88]

Thanks, BvU, I'm starting to think I'm over reaching with this problem and, yes, making too many assumptions. However now I just want to know what the right answer is for myself. Thanks for the insight that g is tangential!

 

[JRMichler]

You might be better off to use numbers that the students can wrap their minds around. Start with a building 100 meters tall, and throw a baseball at 30 meters per second, while emphasizing that "we are ignoring air resistance". Calculate the distance and show with a sketch. Point out that this is roughly the velocity that a good pitcher can throw a baseball.

Then 1000 m/sec for a high velocity rifle for two cases: flat earth, and round earth. Show the trajectory on a diagram with both a flat and curved earth.

Then 7800 m/sec for a science fiction gun. Show the intersection with a flat earth, and how it never hits a round earth.

 

[kuruman]

If your class knows about centripetal acceleration, you can calculate for them the speed $v$ of a satellite at a grazing orbit around the Earth using $a_C=g=v^2/R_E$, where $R_{Earth}$. That should open the door to estimating the time to any point on the surface of the Earth assuming a circular trajectory of an object in free fall.