### Video Transcript

In this lesson on orbital speed,
we’re going to learn how to relate the orbital speed of an object to information
about the gravitational attraction between the object and the body it’s
orbiting.

For an object moving along a
circular path, the velocity of the object always points tangent to the path. And this is true no matter where on
the circle the object is located. The reason the velocity is always
tangent to the path is because the velocity points in the direction of motion. And if we zoom in at any given time
on the portion of the path where the object is located, the path will more and more
closely resemble the line tangent to the path at that point. So, at any given moment, the motion
of the object appears to be along the line tangent to the path. So, the velocity points tangent to
the path at each point.

We’re going to be interested in
objects that have a constant speed as they move around the circle. This means that the size of the
velocity vectors is the same at every point. However, as we can clearly see from
the picture, the direction of the velocity changes. But if an object’s velocity changes
direction, the object is accelerating. We call this acceleration
centripetal acceleration.

As a formula, we can write 𝑎 c,
the centripetal acceleration, is equal to 𝑣 squared over 𝑟, where 𝑣 is the speed
of the object and 𝑟 is the radius of the circular path. This expression gives us the
magnitude of the centripetal acceleration. The direction of the centripetal
acceleration will be radially inward towards the center of the circle. Visually, we can see that this is
the correct direction because in order to keep the velocity vector tangent to the
circle at each point, we must continually move the head of the vector in towards the
center of the circle.

Now, let’s recall the Newton second
law. It tells us that an accelerating
object with a mass of 𝑚 will be experiencing a force whose size is the mass times
the acceleration and the direction is the same as the acceleration. In other words, in order for our
mass to stay on the circular path, it must be constantly experiencing a centripetal
force which has the same direction as the centripetal acceleration. Let’s now bring this discussion to
the specific case of circular orbits that we’re interested in.

In order for our object to stay in
orbit around a planet, there needs to be some source of centripetal force. Well, in a situation like this,
there is one force that acts radially inward between the planet and the object. And that’s the force of
gravity. The direction of the gravitational
force is radially inward. The size of the gravitational force
in terms of the mass of the planet, capital 𝑀, the mass of the object, lowercase
𝑚, and the radius, 𝑟, measured from the center of the planet to the orbital path
is given by. 𝐺, the universal gravitational
constant, times the mass of the planet times the mass of the object divided by the
distance between them squared.

The distance between them is the
same as the orbital radius, since at every point along the path, the object will be
𝑟 away from the center of the planet. Note that when we say the center of
the planet or the center of the object, we mean their centers of gravity. Our stated goal was to relate the
orbital speed of the object to information about the gravitational attraction
between the object and the planet. The last observation we need before
we derive a formula is that the gravitational attraction between the planet and the
object is providing the centripetal force on the object to keep it on its path.

We can, therefore, write 𝐹 c, the
centripetal force, is equal to 𝐹 g, the gravitational force. Let’s now plug in the relevant
information from our three formulas. By Newton’s second law, the
centripetal force is mass times the centripetal acceleration. By using 𝑣 squared over 𝑟 as the
definition of centripetal acceleration, we can write mass times centripetal
acceleration as mass times orbital speed squared over orbital radius. On the other side of the equation,
gravitational force is just given by the expression we mentioned before.

Okay, now, we’re really close to
our goal because we have a term on the left-hand side of the equation with the
orbital speed which we’re looking for and the right-hand side of the equation is
information about the gravitational attraction between the two bodies. We’ll start by noting that there’s
a factor of lowercase 𝑚 on both sides of the equation. There’s also at least one factor of
𝑟 in the denominator on both sides. So, let’s multiply both sides by 𝑟
over lowercase 𝑚.

On both the left-hand side and the
right-hand side, we have 𝑚 divided by 𝑚, which is just one. On the left-hand side, 𝑟 divided
by 𝑟 is one. And on the right-hand side, 𝑟
divided by 𝑟 squared is just one over 𝑟. This leaves us with speed squared
on the left-hand side and universal gravitational constant times mass of the planet
divided by the orbital radius on the right-hand side.

Our final step will be to take the
square root of both sides of this equation. The square root of orbital speed
squared is just orbital speed. And the right-hand side is just as
it was before, square root of 𝐺 capital 𝑀 over 𝑟. We have thus succeeded in finding a
formula for the orbital speed of an object in a circular orbit. In terms of the universal
gravitational constant, the mass of the body that our object is orbiting, and the
radius of the orbit. It’s worth stressing again that
this formula only applies in the special case of a circular orbit where the orbital
speed is constant. If the orbit is not a circle, the
orbital speed will not be constant and this formula will not apply.

Note also that lowercase 𝑚, the
mass of our object, does not appear in this formula. What this means is that all
objects, no matter what their mass is, that are moving in circular orbits with the
same orbital radius around planets with the same mass will all have the same orbital
speed. As we’ve written it, this equation
allows us to find the orbital speed, but we can also rearrange this equation to find
the mass of the planet and the radius of the orbit. We’ll start by squaring both
sides.

This gives us back our previous
equation 𝑣 squared equals 𝐺𝑀 over 𝑟. To solve for 𝑟, we multiply both
sides by 𝑟 over 𝑣 squared. On the left-hand side, 𝑟 over 𝑣
squared times 𝑣 squared is just 𝑟, which is what we were looking for. And on the right-hand side, 𝑟 in
the numerator divided by 𝑟 in the denominator is just one and we’re left with 𝐺𝑀
over 𝑣 squared. In this form, we have orbital
radius in terms of mass of the planet and orbital speed.

If we multiply both sides of our
previous equation by 𝑟 over 𝐺 instead of 𝑟 over 𝑣 squared, then on the
right-hand side 𝑟 over 𝐺 times 𝐺 over 𝑟 is one. And we’re left with just the mass
of the planet which we’ll put on the left-hand side of our final result, just to
keep consistency. And on the left-hand side, we have
𝑟 times 𝑣 squared over 𝐺. And that leaves us with 𝑀 is equal
to 𝑟𝑣 squared over 𝐺. So, with these three equations, we
can find the orbital speed or the orbital radius or the mass of the planet as long
as we know a value for the universal gravitational constant and also have values for
the other two quantities.

Before moving on to a few examples
of how to apply these equations, let’s see one more way to write them. If we solve any of these equations
for 𝐺, we find the universal gravitational constant is equal to the orbital radius
times the orbital speed squared divided by the mass of the planet. What this means is that if we could
independently measure the orbital radius and orbital speed of a circular orbit
around a planet and could also measure that planet’s mass. We could find a value for the
universal gravitational constant that we could then use every other time gravity
comes into play.

We could also test the validity of
our theory of gravity by comparing this measure of the universal gravitational
constant to other measures of the universal gravitational constant. Anyway, let’s now see some examples
of how to apply the formulas we’ve just derived.

For a satellite to follow a
circular orbit around Earth at a radius of 10,000 kilometers, what orbital speed
must it have? Use a value of 5.97 times 10 to
24th kilograms for the mass of Earth and 6.67 times 10 to the negative 11th meters
cubed per kilogram per second squared for the value of the universal gravitational
constant. Give your answer to three
significant figures.

Okay, so here’s the Earth with a
mass of 5.97 times 10 to 24th kilograms. And here is the circular orbit with
a radius of 10,000 kilometers, as measured from the center of the Earth. Finally, here’s a satellite moving
along the orbit with an unknown speed. For the satellite to maintain a
constant circular orbit, it must be experiencing a centripetal force which comes
from the force of gravity that the Earth exerts on the satellite.

Let’s call the mass of the
satellite lowercase 𝑚, which is not a quantity that we’re given in the problem. Using lowercase 𝑚, 𝑣, 𝑟, capital
𝑀, and the universal gravitational constant. We can write 𝐹 c, the centripetal
force, is equal to the mass of the satellite times the square of the orbital speed
divided by the orbital radius. Also, 𝐹 g, the force of gravity
that Earth exerts on the satellite, is equal to the universal gravitational constant
given the symbol capital 𝐺 times the mass of Earth times the mass of the satellite
divided by the square of the orbital radius.

Since the centripetal force is
provided by the gravitational force, we can equate these two quantities and then
solve for the orbital speed. We get 𝑣 is equal to the square
root of 𝐺 times capital 𝑀 divided by 𝑟, whereas we can see the mass of the
satellite does not appear in this final expression. In our question, we’re given a
value for the universal gravitational constant. So combining that with our known
values for orbital radius and mass of Earth, we should be able to plug in to get a
value for the orbital speed.

When we actually put in those
numbers, we find that 𝑣 is equal to the square root of 6.67 times 10 to the
negative 11th meters cubed per kilogram per second squared times 5.97 times 10 to
the 24th kilograms divided by 10,000 kilometers. Let’s start with two
simplifications involving the units. We have a factor of per kilograms
and a factor of kilograms. And per kilograms times kilograms
is just one.

Second, we’ll need to convert
kilometers to meters in the denominator to match the meters already present in the
numerator. Recall that one kilometer is by
definition 1000 meters. So, 10,000 kilometers is 10,000
times 1000 meters or, converting to scientific notation, 10 to the seventh
meters. Finally, meters cubed in the
numerator divided by meters in the denominator is meters squared. Let’s rewrite this expression
separating the numbers, the powers of 10, and the units.

Written out like this, using the
commutativity and associativity of multiplication, we can now calculate the value of
each of these terms separately, take the square roots separately, and multiply all
those together to get the final result. Let’s start with the units. Meters squared per second squared
is meters per second squared. Moving on to the powers of 10, our
final result will have a base of 10. And to get the exponent, we add the
exponents in the numerator and subtract the exponents in the denominator. Negative 11 plus 24 is 13 minus
seven is six. So, that whole term reduces to 10
to the sixth.

Finally, 6.67 times 5.97 is equal
to 39.8199. To finish the calculation, we
recall that the square root of a product of several terms is equal to the product of
the square root of those terms. So, we have that 𝑣 is equal to the
square root of 39.8199 times the square of 10 to the sixth times the square root of
meters per second squared. Okay, so let’s work out these
square roots.

To take the square root of a
squared quantity, well, that’s just the quantity itself. So, the square root of meters per
second squared is just meters per second. This is a good intermediate result
because meters per second is a unit of speed. So, we see that we’re looking for a
speed, and our final answer will have units of speed. In general, to take the square root
of any quantity raised to a power, simply halve the power. So, the square root of 10 to the
sixth is 10 to the third.

Lastly, we have the square root of
39.8199. For this, we just need a
calculator. The first several digits of that
result are 6.3103 et cetera. Now, we have our answer as a number
times the power of 10 times some units, which is a very useful form for expressing
it to three significant figures. To express our answer this way, we
simply need to find the three significant figures of the number portion and then
carry the power of 10 and the units to the final answer.

To identify some number of
significant figures, we count that many digits, starting from the first nonzero
digit and going left to right. For our number, the first
significant figure is six and the second is three. To find the third significant
figure, since that’s the last one that we’re looking for, we have to round. So, we look at the fourth digit,
which is zero. And since zero is less than five,
one rounds to one. So, our number to three significant
figures is 6.31.

Now, let’s just finish out the
multiplication. 6.31 times 10 to the third is 6,310
times meters per second gives us a final answer of 6,310 meters per second to three
significant figures. And this is the orbital speed that
a satellite would need to maintain to have a circular orbit around Earth with a
radius of 10,000 kilometers.

This example was very focused on
calculations. Let’s now see an example that’s
focused on the qualitative relationship between orbital speed and orbital
radius.

Which line on the graph shows the
relation between orbital speed and orbital radius for objects moving along circular
orbits due to gravity?

The question asks us to identify
which of the curves on this graph, which has orbital radius on the horizontal axis
and orbital speed on the vertical axis, corresponds to the correct functional
relationship between those two quantities. Recall that by equating the
centripetal force and the gravitational force for an object in a circular orbit. We find that the orbital speed is
equal to the square root of the universal gravitation constant times the mass of the
body being orbited divided by the radius of the orbit.

Since our question is only about
the relationship between orbital speed and orbital radius, we can treat the mass of
the body being orbited, that is capital 𝑀, as constant. Then, since capital 𝐺 is also
constant, we can rewrite our formula as orbital speed is equal to the square root of
a constant divided by the orbital radius, where we’ve used the fact that a constant
times a constant is just another constant. Now, using the fact that the square
root of a constant is just another constant, we can rewrite this form as orbital
speed is equal to a constant divided by the square root of the orbital radius.

We don’t actually care about the
identity of this constant. So, we can rewrite this equation as
a qualitative proportionality relationship. And that relationship is that 𝑣 is
proportional to one over the square root of 𝑟. We’ve written the relationship this
way to focus on the connection between orbital speed and orbital radius and avoid
getting confused by constants that are not relevant to this particular question. Let’s now use this functional form
to make predictions for the orbital speed corresponding to large orbital radii, that
is the right edge of the graph, and to small orbital radii, that is the left edge of
the graph.

We can then match those predictions
to the appropriate line. Let’s see what happens as 𝑟
increases. As the orbital radius gets larger,
the square root of the orbital radius also gets larger. So one over the square root of the
orbital radius gets smaller because as the denominator of a fraction grows, the size
of the fraction shrinks. So when orbital radius increases,
we expect orbital speed to decrease. Although because orbital radius can
never be infinite, the orbital speed can never be zero.

Looking at the graph, clearly, the
green line doesn’t work because it never changes, so it isn’t decreasing with
increasing radius. The blue, orange, and red lines all
decrease with increasing radius. But it also can’t be the red line
because the red line reaches zero. And we know that the orbital speed
can never be zero. Let’s now see what happens as the
orbital radius shrinks.

As 𝑟 decreases, so does the square
root of 𝑟. So, the denominator of our fraction
is shrinking. And so, the fraction itself is
growing. And since the fraction is
proportional to orbital speed, the orbital speed is also growing. We also know that as the
denominator of a fraction gets closer and closer to zero, the value of that fraction
increases without limit. So here, too, we expect the orbital
speed to increase without limit as the orbital radius gets closer to zero. The orange curve has a maximum
value as the radius approaches zero. So, the orange curve is not the
right curve because the orbital speed doesn’t increase without limit.

This leaves the blue line as the
correct answer. Indeed, the blue line shows an
orbital speed that gets smaller and smaller as the radius increases and gets larger
and larger without limit as the radius decreases. So, the answer is that the blue
line shows the correct relationship between orbital speed and orbital radius for
objects moving in circular orbits due to gravity.

Now that we’ve seen some examples,
let’s summarize the key points that we’ve learned in this lesson. We can find our discussion to
circular orbits caused by gravitational fields. In such a case, we found that we
could equate the centripetal force, keeping the object moving in a circle, to the
gravitational force acting on the object.

Using the definition of centripetal
force as mass of the object times speed squared divided by the radius. And the definition of gravitational
force as the universal gravitational constant times the mass of the planet times the
mass of the orbiting object divided by radius squared. We derive the equation for the
orbital speed as 𝑣 is equal to the square root of the universal gravitational
constant, 𝐺, times the mass of the planet, capital 𝑀, divided by the orbital
radius, 𝑟.

We then solve this equation for 𝑀
and 𝑟 to get two more formulas. We found that the orbital radius is
equal to 𝐺 times the mass of the planet divided by the orbital speed squared and
that the mass of the planet is equal to the orbital radius times the orbital speed
squared divided by 𝐺. Finally, we noted that given an
appropriate value, for 𝐺, we could find orbital speed, orbital radius, or mass of
the planet from values of the other two quantities.