WEBVTT
00:00:03.300 --> 00:00:06.420
Let's consider the universal
gravitational law a little bit
00:00:06.420 --> 00:00:08.950
more.
00:00:08.950 --> 00:00:11.340
Let's consider two
objects in space.
00:00:11.340 --> 00:00:15.120
Let's say this is the sun,
and we have the earth here
00:00:15.120 --> 00:00:17.430
or the earth and the moon.
00:00:17.430 --> 00:00:19.590
And of course, they're
orbiting each other.
00:00:19.590 --> 00:00:23.430
So we can pick a coordinate
system that goes radially.
00:00:23.430 --> 00:00:26.485
And so we're going to have
an r hat direction here,
00:00:26.485 --> 00:00:30.400
and we're going to call this the
r hat direction between objects
00:00:30.400 --> 00:00:33.570
1 and 2.
00:00:33.570 --> 00:00:36.370
What forces are acting
on this little moon here?
00:00:36.370 --> 00:00:40.520
Well, it's the gravitational
force going inward.
00:00:40.520 --> 00:00:44.750
It's force of 1, 2,
F1 2, on object 2,
00:00:44.750 --> 00:00:48.590
due to the interaction
between bodies 1 and 2.
00:00:48.590 --> 00:00:52.620
For that, we can write down the
universal gravitational law,
00:00:52.620 --> 00:01:01.060
F1 2 equals minus G, the
gravitational constant, m1,
00:01:01.060 --> 00:01:05.280
m2 of r1 2 squared.
00:01:05.280 --> 00:01:12.190
That one is the distance between
the two objects times r1 2 hat.
00:01:12.190 --> 00:01:14.660
And the minus goes, actually,
with the unit vector
00:01:14.660 --> 00:01:17.440
here, because the force goes
in the opposite direction
00:01:17.440 --> 00:01:20.270
from our r hat.
00:01:20.270 --> 00:01:23.789
Let's now consider here
is the earth again.
00:01:23.789 --> 00:01:28.200
And we're going to move the
moon or a little moon rock right
00:01:28.200 --> 00:01:29.780
to the surface of the earth.
00:01:29.780 --> 00:01:33.800
And we want to now calculate
and consider what kind of force
00:01:33.800 --> 00:01:36.259
this act on this
moon rock, and what
00:01:36.259 --> 00:01:38.820
is the gravitational
acceleration that this moon
00:01:38.820 --> 00:01:41.850
rock on the surface of
the earth is experiencing?
00:01:41.850 --> 00:01:44.110
So we have the earth.
00:01:44.110 --> 00:01:48.680
Earth has one earth radius,
and it has an earth mass.
00:01:48.680 --> 00:01:53.880
And our moon rock
has the mass m.
00:01:53.880 --> 00:01:58.090
And we know, from this exercise
here already, that, of course,
00:01:58.090 --> 00:02:01.620
this gravitational force
is acting on our moon rock
00:02:01.620 --> 00:02:02.990
as well.
00:02:02.990 --> 00:02:04.510
That hasn't changed.
00:02:04.510 --> 00:02:06.610
What we are now
considering in addition
00:02:06.610 --> 00:02:09.038
is that this moon rock
is also experiencing
00:02:09.038 --> 00:02:12.300
a gravitational acceleration
due to this force,
00:02:12.300 --> 00:02:13.960
and that goes inward as well.
00:02:13.960 --> 00:02:18.090
So it is experiencing an mg.
00:02:18.090 --> 00:02:23.360
And we know that that is
the same as the magnitude
00:02:23.360 --> 00:02:25.520
of this force here.
00:02:25.520 --> 00:02:31.920
So we can equate that with G,
and then we have the earth mass
00:02:31.920 --> 00:02:36.440
and the mass of the moon rock
times the distance squared,
00:02:36.440 --> 00:02:40.220
so an earth radius squared.
00:02:40.220 --> 00:02:43.840
And from that, we
already see that a, we
00:02:43.840 --> 00:02:47.500
can cancel out the small
m, so the moon rock,
00:02:47.500 --> 00:02:49.170
and we get to g here.
00:02:49.170 --> 00:02:53.360
So we can calculate the
gravitational acceleration,
00:02:53.360 --> 00:03:00.820
which is capital G earth mass
over earth radius squared.
00:03:00.820 --> 00:03:03.150
So if we have this
kind of information,
00:03:03.150 --> 00:03:05.840
we can determine the
gravitational acceleration.
00:03:05.840 --> 00:03:08.690
And of course, it will
change, depending on which
00:03:08.690 --> 00:03:10.800
object we are considering.
00:03:10.800 --> 00:03:14.120
It would be different if
we plug in the solar mass
00:03:14.120 --> 00:03:17.350
and the solar radius or the
moon mass and the moon radius,
00:03:17.350 --> 00:03:20.760
if we consider an
astronaut standing here
00:03:20.760 --> 00:03:23.980
on the moon's surface.
00:03:23.980 --> 00:03:28.079
Now let's put some numbers
into this equation.
00:03:28.079 --> 00:03:33.470
So we have g is
capital G. That's
00:03:33.470 --> 00:03:34.860
the gravitational constant.
00:03:34.860 --> 00:03:43.210
We have 6.67, 10 to 11, and
then we have Newton and 1
00:03:43.210 --> 00:03:49.980
over kilogram squared
and mass squared times
00:03:49.980 --> 00:03:57.650
the earth mass, 5.97
10 to 24 kilograms.
00:03:57.650 --> 00:04:01.150
And then we have
to divide this over
00:04:01.150 --> 00:04:07.730
through the earth
radius, 6.37 10 to the 6.
00:04:07.730 --> 00:04:12.311
And we have to square that, and
we have to square the meters.
00:04:14.960 --> 00:04:21.959
If we calculate this, we get to
9.81 meter per second squared.
00:04:21.959 --> 00:04:25.580
And surely you have
seen this number before.
00:04:25.580 --> 00:04:27.180
This number can
either be calculated,
00:04:27.180 --> 00:04:30.490
if you know capital G, the
gravitational constant,
00:04:30.490 --> 00:04:33.930
or you can determine that
gravitational acceleration
00:04:33.930 --> 00:04:35.670
through an experiment.
00:04:35.670 --> 00:04:38.710
And together with the earth
mass and the earth radius,
00:04:38.710 --> 00:04:43.380
you can actually calculate the
gravitational constant there.