Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[wierd]

You wouldnt be able to get closer.  It would be radioactive as hell. It would also have an EVEN STRONGER magnetic field. The geyser plumes from io, europa and pals would get pulled into a thin accretion disc around the object, and emit super deadly hard Xrays, and be absurdly energetic electromagnetically.

[MonkeyHead]

It would be hella bright as a result - if time allows this weekend, I might go ahead and calculate just how bright.

[10ebbor10]

You wouldnt be able to get closer.  It would be radioactive as hell. It would also have an EVEN STRONGER magnetic field. The geyser plumes from io, europa and pals would get pulled into a thin accretion disc around the object, and emit super deadly hard Xrays, and be absurdly energetic electromagnetically.

Not really. The plumes are emmited from the moons, and thanks to conservation of momentum would roughly keep their orbit. Unless something disturbs them, they won't fall into the hole.

[wierd]

While the mass is the same, the local curvature of spacetime wont be.

By definition, a singularity is where spacetime curvature is infinite. The characteristics of the gravity well will be different. The magnetosphere of the system will be different.

ergo, the paths of gas particles in the system will be different.

[10ebbor10]

The mass is the same, so the gravitational effects are the same, especially at that range. After all, all mass disturbs spacetime. The difference noticed will be much closer than the core than most particles are.

Magnetosphere might get complicated though.

[Cthulufaic]

Alright, new question, just cause this one has been bothering me for a while:  How much force would be exerted on a basketball hoop if someone slam dunked it from 300ft in the air(straight up if it simplifies things)  This is assuming the slam dunk'er weighs 200kg and the basketball is 5kg and on Earth*

*Bonus challenge: On Mars.

[wierd]

Insufficient.

What elevation is the basketball hoop. G falls off in the inverse square of distance from center of mass, and G is needed to calculate accelleration.

We need to know the distance the basketball hoop is from the center of the planet. (Most calculations assume sea-level, but this is not accurate!)

[Cthulufaic]

The hoop is 15ft off the ground, which is at 1000 feet above sea level.  Assume constant air resistance(don't really know how to phrase this, still in chemistry, next year is physics)

[wierd]

That is not what I need.

I need either of these two things:

Elevation relative to sea level (exercise of solver to determine radial distance from planetary center of mass)

Radial distance from planetary center of mass.


The bonus question, "on mars", requires the radial distance, because mars has no ocean.

You can make this more challenging by asking "what will the force on the basketball hoop be in Jakarta?" (A city who's mean elevation above sea level can be obtained, and from which the radial distance to the earth's core can be derived) and bonus "On Planitia Marinaris?"-- a geological feature on the surface of mars, that has a known elevation relative to the planet's center.

[Cthulufaic]

Check the edit, just made it.  The hoop is 1015ft above sea level.  Also, for the Mars bonus, the hoop is at at the tallest point on Olympus Mons.

[wierd]

YAY.

I don't have the time to actually work the problem at the moment though.

If nobody answers before tomorrow morning, I will do it then.

[Il Palazzo]

This is silly. You don't need to know all those things, as the distances involved are too tiny a fraction of the planetary radius. Just treat gravity as a constant and you'll get a very good approximation of the actual answer.
Also, disregard the air resistance, unless you feel unhealthy excitement whenever you see a differential equation.

Use mgh to find PE at 100 metres, convert to KE, find the velocity(~45m/s), then the momentum(~9000 m kg).

Assume the hoop is able to completely stop the player, so the force will be Δp/Δt under the(unrealistic) assumption of constant force over time.
You now just need to find the time frame in which the hoop does the stopping.
I don't know how much the hoops can normally bend, so let's say it's 20cm displacement in the vertical direction, over which distance the velocity goes down from 45 to 0 m/s. Average V=22.5 m/s, at which speed covering the 20cm distance takes 0.009 seconds, and the force required is in the vicinity of a million newtons.

Unless the hoop can survive having a 100 metric ton dead weight hung from it, it will break during the slam dunk.

Edit:
That's on Earth. On mars you get V=~27m/s and p=~5500mkg, so F=~370 kN, or again, a 100 metric ton dead weight.

[wierd]

[Elevation doesn't impose a significant difference]
[Redefines problem to avoid differential equasion where elevation plays a significant role]
[Gives answer to redefined problem]


Honestly. The air pressure and viscosity changes considerably with elevation. So, yes, YOU DO need that to answer the question as asked. Redefining the question is cheating!

[Il Palazzo]

Go ahead and calculate it your way and see if it was worth the extra fine precision.
And it's not cheating, it's finding the upper bound. Or solving a Fermi problem. Or being an astronomer.

[misko27]

I'd be able to do it but the air resistance throws me off.

Ninjaedit: Wow I calculated energy instead. Well, here you can have this if you want

Spoiler (click to show/hide)

Well, PE=mgh, so let's do that. m is easy, 205 kg; h is only slightly more difficult in conversion, 91.44 m. g is hard.

g=GM/r^2. G is universal gravitational constant, 6.67x10^-11. M is mass of earth (mass of basketball player is irrelevant at this scale), at 5.976x10^24. r is radius from gravitational mass, which is 3,959, plus 705.612 for height differences. This lands us a healthy 1.8593106x10^18.

Plugging that in gives us 3.4853149x10^22 J, assuming no air resistance. With Air resistance, I have no idea. I might be off in any case.