Research Interests:
Application of imaging spectroscopy and light detection and ranging (lidar) for spectral-structural characterization of natural systems (remote sensing of natural resources)
Forestry (vegetation) inventory and assessment
Species diversity
Foliar biochemistry
Scaling of remote sensing estimates
Escape velocity and black holes
We have seen that one can compute the escape velocity
from a body via
Lets take a massive object -- like the Earth --
and squeeze it down into a smaller size.
1. Suppose that Earth's radius were only R = 60 km.
What would its escape velocity be?
2. Suppose that Earth's radius were only R = 150 m.
What would its escape velocity be?
3. Suppose that Earth's radius were only R = 10 cm.
What would its escape velocity be?
If we keep squeezing and shrinking the Earth,
we eventually reach a critical point:
the escape velocity reaches
c, the speed of light.
4. How small would the Earth have to be for
its escape velocity to be the speed of light?
If an object somehow becomes this small,
then light rays from its surface cannot get away.
The object would become .... black.
We call such extremely dense objects
black holes, because
they do not directly emit any light
their gravitational force attracts objects,
which can fall into them
Ordinary objects do not, and can not, turn into black holes.
However, certain types of stars can and do!
For example, a star which starts its life with
about 20 times the Sun's mass will eventually run out of
nuclear fuel at its core;
when that happens, the core will collapse into a black hole
(yes, I'm simplifying :-)
5. What would be the radius of such a black hole,
with a mass 20 times the Sun's?
Tidal forces
One reason that black holes CAN be dangerous is simply
because they are very small,
and therefore, it is possibly to approach one very close.
If one comes very close to a massive object,
then the differences in its gravitational force on
different parts of your body can become important ...
or even painful.
Let's see how that works:
6. Joe hangs by his hands from a bar.
The Earth, with radius
M = 5.98 x 10^(24) kg
and radius
R = 6,370,000 m,
pulls on all parts of his body.
Since his feet are a bit closer to the center
of the Earth than his head, the gravitational
force is stronger on his feet than his head.
Assume that Joe's feet and head both have
mass m = 5 kg,
and that Joe's height is h = 2 m.
What is the gravitational force on Joe's head?
Don't round off!
What is the gravitational force on Joe's feet?
Don't round off!
What is the difference between the gravitational
forces on Joe's head and feet?
Express this difference as a percentage of
the force on Joe's head.
Now, suppose we change the arrangement a bit:
we take Joe to planet X, which is smaller
than Earth.
Planet X has
M = 1.47 x 10^(17) kg
and radius
R = 1000 m.
What is the gravitational force on Joe's head?
Don't round off!
What is the gravitational force on Joe's feet?
Don't round off!
What is the difference between the gravitational
forces on Joe's head and feet?
Express this difference as a percentage of
the force on Joe's head.
We change it one more time:
Joe hangs above a tiny sphere of very dense material.
The sphere has
M = 1.47 x 10^(13) kg
and radius
R = 10 m.
What is the gravitational force on Joe's head?
Don't round off!
What is the gravitational force on Joe's feet?
Don't round off!
What is the difference between the gravitational
forces on Joe's head and feet?
Express this difference as a percentage of
the force on Joe's head.
The magnitude of the tidal forces exerted
by an object depends on the
change in gravitational force as one moves
away from the object:
if the gravitational force changes quickly,
then the tidal forces are large.
Mathematically speaking,
we can say
6. As you travel towards a black hole,
you notice a strain of about 50 N
between your head and feet.
You quickly manuever your spaceship
to double the distance between you
and the black hole.
How big is the strain between your
head and feet now?
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