Question

A discussion on astroids led me to ask this question.
How much power would a tungsten mass with about the size of a volleyball do to the earth and terminal velocity and about melting temperature?

Answer 1

After a bit of calculation, it seems that a volleyball has a radius of about 5 inches.

Answer 2

Density near room temp:19.25 g·cm^(−3)
Density near melting point:17.6 g·cm^(−3)

Answer 3

Now, we will assume that the mass of tungsten starts out at room temperature, and we can safely ignore the densitry near melting point, until we decide to use it to calculate the heat energy that the ball has near melting point.

Answer 4

Now the volume of a sphere is \(V=\pi*r^3\)

Answer 5

we have \(M=VD\)

Answer 6

or, \(M=\pi*5^3*19.25\)

Answer 7

Because we are making a rough estimate, I will use rounding. First I will round pi to 3, and 19.25 to 20.

Answer 8

M approximately equals 3*(13)^3*20, or NOTE THAT I HAVE CONVERTED TO CENTIMETERS AND ROUNDED.

Answer 9

1683240 is our approximate answer in grams here, which is around 1680 kg.

Answer 10

Now, what's the equation for kinetic energy?

Answer 11

E=(1/2)mv^2

Answer 12

so, we have our mass; let's find the terminal velocity. I shall google this, because it is a very bad idea to mess around with calculus.

Answer 13

The projected area of a sphere is pir^2 according to wikipedia, and wolfram alpha helps us calculuate the terminal velocity here.

Answer 14

I got a lot of things wrong, give me a second to fix them.

Answer 15

http://www.wolframalpha.com/input/?i=terminal+velocity&a=*C.terminal+velocity-_*Formula.dflt-&f2=100000+ft&f=TimeToFall.h_100000+ft&f3=100000+ft&f=TimeToFall.H_100000+ft&f4=air&x=10&y=11&f=TimeToFall.rho_air&f5=165kg&f=TimeToFall.m_165kg&f6=0.47&f=TimeToFall.Cd_0.47&f7=%28pi*%2812.7^2%29%29%2F1000&f=TimeToFall.A_%28pi*%2812.7^2%29%29%2F1000&a=*FVarOpt.1-_***TimeToFall.H-.*TimeToFall.h-.*TimeToFall.m-.*TimeToFall.Cd-.*TimeToFall.rho-.*TimeToFall.A-.*TimeToFall.withDrag--.***TimeToFall.d---.*--

Answer 16

i.e.
Velocity=480
Mass=165 kilograms

Answer 17

It's very close to 1kilowatt hour.

Answer 18

about 0.4 times the power of a general car battery. You must realize that this energy is displaced in a very short time, maybe 0.1 seconds; in other words, you would get about 3600 gigajoules in about 0.1 second.

Answer 19

We have assumed drag for this case. What if we had no drag for this ball of tungsten?
In this case, we must specify a height, because there would be no terminal velocity. If our ball was dropped at 100k feet (around 20 miles)

Answer 20

@FoolForMath , please help. I am unsure of how to calculate the potential energy of an object with a non-constant gravitational pull.

Answer 21

asteroids or meteorites are much faster