You may have read of interplanetary
probes such as the Venera and Pioneer missions to Venus, the Mariner flyby of
Mercury, the Viking excursions to Mars, or the famous Voyager Grand Tour of
the outer planets. In any case, you know of the Apollo series that led to Lunar
landings and exploration.
To launch such a probe, we
have to calculate carefully the velocity needed for the probe to leave the Earth.
This is called "escape velocity", and will be explained below. Next,
we have to determine the trajectory of the target planet by determining its
orbital speed. Then we must decide how to "aim" the probe so that
it will arrive at its destination at the same time as the target planet.
Suppose our task is to determine
the velocity required for a probe to "escape" the Earth's own gravitational
pull. We take the mass of the Earth to be 
.
And we assume that it has radius 6371 Km. Then according to Newton's Universal
Law, the gravitational acceleration that pulls the rocket back to Earth is,
at height h Km above the surface of the Earth,
We will use the number 
for 
, and we measure distance to center of the Earth in Km. We will assume that
the probe will fly radially away from the center of the Earth, and will later
"take aim" with some directional course adjustment.
According to Newton, at any
point in time, the 
of the probe will be the sum:
|
|
This is KineticEnergy+PotentialEnergy divided by the mass.
We will ignore the decrease
in mass of the probe that is due to the loss of fuel once the probe is launched,
the quantity above will not vary in time until the probe comes into the vicinity
of the target planet. We can safely ignore the gravitational effect of the target
for now.
Now this is one important
place where Newton's story diverges from Galileo's. The fact that this quantity
is conserved (is constant) means that it must be the same at the moment the
engines stop firing as it is for the rest of its trip.
At that point, its vertical
speed will determine whether or not is has enough "energy" to escape
the Earth's gravitational pull. How is this? According to Newton, the acceleration
pulling the probe back decreases with distance since it is inversely proportional
to the square of the distance. Not so with Galileo. For Galileo, that acceleration
is constant.
Thus, if the probe is moving
fast enough, it may "outrun" the Earth's gravitational pull in a sense.
It may be always far enough away that the acceleration pulling it back is too
weak to turn it around. If that happens, we will say it "escapes".
Of course it always feels some acceleration back to Earth. But that may be too
weak to be effective.
Question 1: Given that energy/mass is constant, determine the speed the probe must have, if at the moment the engines shut down (given that the expression below is constant)
the speed cannot later become 0. (We assume here that the engines shut down immediately, and like in the Jules Verne novel, the rocket is "kicked" into space. This is not the way things actually happen, but the principal is the same, it is just that the initial height is in reality the height at which the engines do in fact turn off.) Take this initial radius to be 6371 Km. The speed you find will be the escape velocity. To do this, write an equation that asserts that this expression at time 0 is equal to the expression at any later time and see if there are values for the speed at time 0 that make it impossible for the speed later to be 0. Take the smallest of these values. Use the calculator page if you want to.
On this page, you may experiment
with these two pictures: Galileo's picture, and Newton's picture of gravitation.
Simply set the initial velocity of the probe with the scroll bar:
And press one of the "Launch Buttons". You may stop the simulation at any time by pressing the P key or (Esc). The graphics windows both represent time on the horizontal axis (in seconds) and position on the vertical axis (in Km). The range of variation of position is from 0 to 10000 Km. The range of time is from 0 to 5000 sec. Velocity is measured in Km/sec.
The graphs will show position
portraits and velocity portraits superimposed, as on an earlier page. For convenience,
the unit vertical interval represents 1 Km/sec (not 1000 Km/sec), so that you
can see changes in the velocity.
Question 2: According to Galileo: What goes up must come down. We saw that the curve becomes "flat" (We say it develops a horizontal tangent line, or a zero derivative with respect to time) when the velocity becomes 0. Can you guess the relationship between the amount of time the rocket is in the air, and amount of time that passes until the velocity becomes zero? Try to prove this from your formula for the height, and from general properties of parabolas.
Question 3: In Question 1, you were asked to calculate the escape velocity. This Microworld cannot determine it, or even demonstrate that it exists! Later, we will see how to prove it exists. But look at the pictures, and see if your calculated number is plausible. Notice that in Newton's picture, the velocity portrait is not a straight line but is curved! Is it also true here that, if it becomes 0 the rocket turns around? Must it become 0?
Question 4: This is not to be taken too seriously. Try for a while to write a function whose graph is the position portrait. Do you think it is a parabola as in Galileo's case? Why not? How might you determine, then, the time when the probe turns around (if it turns around)? And how might the height it ascends to be determined? Hint: Pay attention to the velocity portrait.
Question 5: Assuming the probe "escapes", must the velocity approach 0 as the time goes on? This is tricky, and requires a close look at Question 1 again. If these exercises appear difficult, it is because some of them are. Some of them cannot be done without the Calculus that we shall develop later. Question 4 really requires the use of differential equations. These are all part of the same body of ideas. The Calculus was invented, really, to describe Gravitation. It happens that it also describes a great many other things.
