And our next problem is problem 3.7.
3.7 deals with circular orbits and
satellites and orbital periods.
It's always a little tricky, always a little tricky.
Here we have the earth and the earth has a radius r earth.
I will use this symbol for earth.
And v earth at a height h above the
surface over the earth--
that is here height h.
At a radius little r.
So h equals little r minus capital R. There's a satellite
going around the earth.
Here is the satellite and here is its velocity.
And let's assume that it's a circular orbit, so the
magnitude of v is constant.
Clearly there will be a force pointing exactly towards the
center of the earth.
Let's call that force F. Let us call the gravitational
acceleration at the surface of the earth, let's call that g
with this symbol for the earth so that there's never any
confusion what I have in mind.
Oh, I should have put the symbol here too.
Well, at location r, the gravitational acceleration is
less because I'm farther away from the center of the earth.
And the gravitational acceleration just like the
gravitational force goes as inverse r squared.
And so this value equals the value that you would have
measured right at the surface of the earth times R at the
earth squared divided by r squared.
And notice, since r is larger than this capital R, this
value is smaller than the gravitational
acceleration at earth.
Now the force that I have here, this gravitational pull
equals m times this gr.
But it also equals m v squared over r, which is the
centripetal force, which is required for this object with
mass m to go around in a circular orbit.
Now I can replace this v by the circumference 2 pi r
divided by the time to go around g.
This ins now the period in seconds to go around.
And I can square that.
Now, the m's cancel.
And if I combine this part of the equation with that part of
the equation, then I find that gr is this quantity.
Oh, by the way, I forgot a little r here.
This r must be here.
So gr equals this quantity, but I can
replace this gr by this.
And so I'm going to get g measured at the surface of the
earth times the radius of the earth squared
divided by r squared.
That now equals to 4 pi squared.
This r sub 1 r here, so I end up with only one r
divided by T squared.
And out pops the orbital period of this satellite,
which equals 2 pi times the square root of r cubed divided
by the radius of the earth squared times the
gravitational acceleration as measured at the earth.
And what you see here now, perhaps a little bit to your
surprise, that this equation doesn't contain the mass of
the earth and it doesn't contain the gravitational
constant g.
I'll get back to that in a minute.
When you look at this equation, there are a few
things quite intuitive.
Notice that the larger r is, the larger T is.
So if r is larger-- that's my shorthand notation--
T is larger.
Well we all know that the orbital period of the moon
around the earth is substantially larger than the
orbital period, for instance, of the shuttle.
The orbital period of the moon around the earth, if I round
that off is about 25 days.
And the orbital period of the shuttle is
something like 90 minutes.
So it's completely pleasing that r is upstairs here and
when r goes up that T goes up.
Of course, it's by no means obvious that this one is
proportional to r to the power 3/2.
That's is a little bit harder to see to say the least.
But in any case, qualitatively, this goes
exactly along the lines of our intuition.