| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 430.1 | | CORVUS::THALLER | | Thu Jan 16 1986 16:54 | 30 |
| Your argument is flawed. When you state the amount of energy required to
reach height h, you do not take into account that the gravitational pull
reduces as your distance increases (should use an integral). The correct
answer should be something like:
The amount of work required to move a man from the surface of the asteroid
to infinity is
kMm(r^2)/R^3
This is usually expressed in escape velocity or
v = sqrt(2kMr^2/R^3) note that this is independent of mass
Let's assume that our man can jump 1 meter high on earth. This implies
that he can exert energy
E = (1/2)mv^2 = (1/2)(m)(2as) = (9.8)(m) 9.8 is acceleration on Earth
Equating to the energy required for escape velocity and solving for r, you
get:
r = sqr[(9.8)(R^3)/(MK)]
substituting for the constants, (man's mass = 80kg)
r = sqr[(9.8)(6.4E6)^3)/(6.0E24)(6.67E-11)] = 2500 meters or about
a radius of 1.6 miles. Happy Hunting, I hope you find one.
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| 430.2 | | MENTOR::KOSTAS | | Thu Jan 16 1986 20:59 | 21 |
| re>.1
My arguments are correct, and you are partially correct, when
stating that should use an integral. I will give an additional hint:
+
At a distance x from the center of the asteroid, it exerts on the
man a force equal to km(r^3/R^3)M/x^2, and the energy required to
raise him to a height = H above its center is:
(integral goes here )
. . . .
-
If this hint does not give the answer I will post the correct solution
in three weeks or less.
Enjoy.
Kostas G.
<><><><><>
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| 430.3 | | TURTLE::GILBERT | | Mon Jan 27 1986 18:14 | 5 |
| How far off the asteroid must he be before he's considered permanently off?
I envision our man jumping 3000 miles, only to be slowly and surely drawn
back to that little dirty icy gravity well (actually, both the asteroid
and he would fall toward each other).
|
| 430.4 | | GALLO::APPELLOF | | Tue Jan 28 1986 08:38 | 3 |
| Way off. I think we're talking escape velocity here. That means he
will never again visit this particular ball of slime.
|
| 430.5 | | GALLO::APPELLOF | | Tue Jan 28 1986 09:19 | 10 |
| The answer in .1 was correct. It doesn't matter what the man's mass is.
If he can jump to a height h on Earth, then the asteroid's radius is:
r = sqrt( h * R) where R is the radius of the earth.
For us mere mortals who can only jump 1 meter on Earth, that translates
into an asteroid of about 2500 meters radius. For men of more olympian
capabilities, larger asteroids are possible.
On such an asteroid, there is one cardinal rule: DON'T SNEEZE :-)
|
| 430.6 | | 57557::FONSECA | | Wed Jan 29 1986 13:06 | 11 |
| Jumping 1m on earth does not mean your center of gravity reached
a point 1m above your resting center of gravity. Even when I was
in high school and in good shape, I couldn't touch much above a
point 2 ft. above my resting reach.
To maximize the jumper's efforts he would probably want to take a
running jump and leave at a tangent to the surface.
Or maybe enter a highly eliptical orbit where his body is within
kicking distance every orbit, and slowly build up the necessary
energy :-)
|
| 430.7 | | KEEPER::KOSTAS | | Wed Jan 29 1986 16:43 | 55 |
| Well,
Here is one correct solution by an anonymous professor, to the problem:
"How small would an asteroid be, from which a man could jump permanently off?."
If the asteroid is too small, its downward pull does not use up his kinetic
energy, and bring him to rest, in time for him to be pulled back, and so he
might jump clear off his little world.
Suppose the asteroid has radius = r, while the Earth has radius = R, and that
they are both spherical, and have the same average density. M is the mass of
the Earth, (r^3/R^3)M is the mass of the asteroid, and k is the gravitation
constant. The men's mass = m, and at the surface of the Earth he is attracted
with force = kmM/R^2. In order to jump to a height = h, he must overcome this
force through a distance, and so in jumping must be able to expend energy =
h*kmM/R^2. at a distance x from the center of the asteroid, it exerts on the
man a force = km(r^3/R^3)M/x^2, and the energy required to raise him to a
height = H above its center is the integral:
/ x=H
|
| km(r^3/R^3)M/x^2 (-dx) = km(r^3/R^3)M(1/r-1/H)
|
/ x=r
If he raises forever, 1/H = zero. Equating the energy required for this case
to the jumping energy that carries him to a height = h or Earth, we have
hkmM/R^2 = km(R^3/R^3)M/r
hence
h : r = r : R
The radius of the asteroid is a mean proportional between the radius of the
Earth and the height he can jump on Earth. By h, the height to which a man
can jump, is meant the height to which he can raise his center of gravity,
not the height of a bar he can clear by raising his legs. For the sake of
an example, let us suppose that this height is 4 feet. We shall use 4000 for R,
so we must take h in miles also, so we use h = 4/5280 then
r^2 = 4000 * 4 /5280 = 100/33 = 3.030
This makes the diameter of the asteroid close to 3.5 miles.
The smaller asteroids are now thought to be very irregular in shape, they
are like chunks of shattered globe, we have assumed this one to be spherical
for calculation convenience. To get an idea of what such a sphere is like,
consider a 6 foot man, weighing 180 pounds, he would be 3 miles away from his
horizon. On the asteroid he would weight a little over two ounces, and could
see only 375 feet of his horizon. The entire area of the asteroid would be
37.5 square miles, and our man, walking around its equator, about 11 miles,
would see about one-twenty-fifth of the whole surface.
Enjoy,
Kostas G.
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| 430.8 | | SIERRA::OSMAN | | Sat Feb 01 1986 15:51 | 17 |
| Is there REALLY such thing as escape velocity ? I mean, once you've jumped
off something, your jump is no longer giving you any force, only an initial
velocity.
The only force on you, ignoring the rest of the universe (!), is the
gravitational force pulling you and the asteroid back together.
Hence it seems to me that with enough time, no matter how hard you jump,
you and the asteroid will EVENTUALLY fall back together, or perhaps enter
an orbital relationship (perhaps this latter is considered "escape velocity"?).
Maybe the flaw in my reasoning is that for sufficiently large velocities,
the decrease in gravitational force is going to recede with distance faster
than the resultant slowing down, and hence although there will indeed always
be a gravitational force, the velocity, although decreasing will always be
"greater" than the fighting force ?
/Eric
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| 430.9 | | TLE::BRETT | | Sat Feb 01 1986 19:37 | 16 |
| If you integrate the force on an object as it approaches another you get
the amount of work done = potential energy lost. When the kinetic energy
of the jump is greater than this you do indeed jump clean out of the
gravitational well.
This result holds for the gravitational field around an asteroid...
Now, consider this instead...
I have an infinitely long piece of straight wire, that has a slight
positive charge. Can I throw an electron away from this wire at a sufficient
velocity so it will never return (ie: is there an "escape" velocity) ?
/Bevin
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| 430.10 | | R2ME2::GILBERT | | Sun Feb 02 1986 00:40 | 1 |
| Bevin, are you implying that the universe might never collapse? :-)
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| 430.11 | | BEING::POSTPISCHIL | | Sun Feb 02 1986 11:25 | 15 |
| Re .9:
No.
Along these lines, an interesting thing to do is get a graphics display and
write a program to plot the paths of two or more bodies in free fall. Besides
the obvious parameters of mass, initial position, initial velocity, and time
resolution, one more parameter should be the dimension of space. With two
bodies in three dimensions, you will see only conic sections, but the other
dimensions have an interesting variety of paths, even with just two bodies.
For example, one body might spiral in and then out around another. Don't
forget to try fractional dimensions, also.
-- edp
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