Video Transcript
Two objects, object A with a mass
of five kilograms and object B with a mass of 100 kilograms, are near an even larger
object with a mass of 10 to the 20th kilograms. Object A and object B are at an
equal distance away, 100 kilometers, from the center of mass of the very large
object. What is the magnitude of the
gravitational force experienced by object A due to the very large object? Give your answer to three
significant figures.
Okay, looking over at our sketch,
we see a five-kilogram object. That must be object A. And as well, we see a 100-kilogram
mass. This must be object B. Object A and object B interact
gravitationally with this very large spherical object. Now, this object is represented
here using a rectangle that’s not very much bigger than A and B. But we can imagine that actually
this object is so large it couldn’t fit on the screen at this scale. We have an indication of this
thanks to the very large mass of this object.
So, we could almost think of these
three objects, A and B and our large spherical one, as though our large spherical
object is the Earth, object B, say, is a person, and object A is a book the person
is holding on to. That’s roughly the scale we’re
talking about as we consider these three objects. Our first question related to this
scenario says, what is the magnitude of the gravitational force that object A
experiences due to the very large object?
As we get started figuring this
out, let’s record some of the information we’ve been given. First, we’ve been told that our
large spherical object has a mass, we’ll call it capital 𝑀, of 10 to the 20th
kilograms. And we’re also told that the
distance between the center of mass of this large spherical object, that is, where
all of its mass is effectively concentrated, and our two other objects, object A and
object B, is the same. It’s 100 kilometers. We’ll call that distance 𝑟. And as we see, it’s the same for
object A and object B in relation to our large spherical object. With these values written down,
let’s clear some space and consider again our question.
We want to know the magnitude of
the gravitational force experienced by object A, that’s this object here, due to our
very large spherical object. So, let’s recall the general
equation, called Newton’s law of gravitation, that describes the force of gravity
between two objects that have mass. This equation tells us that that
force is equal to the product of those masses. We’ll call them capital and
lowercase 𝑚, respectively. Multiplied by the universal
gravitational constant, capital 𝐺, divided by the distance between our two masses
squared.
Now, the gravitational constant,
big 𝐺, is approximately equal to 6.67 times 10 to the negative 11th cubic meters
per kilogram second squared. So, if we want to know the
gravitational force on object A caused by the very large object, we can call this
force 𝐹 sub A, then we’ll take this constant and we’ll multiply it by the mass of
our large object. And multiply that by the mass of
object A, we’ll call it 𝑚 sub A, and divide all this by the distance between the
centers of mass of object A and our large spherical object squared.
When we substitute in the values
for all these variables, we’re just about ready to calculate this force magnitude,
except for one thing. Notice that the distance in our
denominator is in units of kilometers. In order to agree with the distance
units in the rest of our expression, we’d like to convert this into meters. We can recall that one kilometer is
equal to 1000 meters and therefore 100 kilometers is equal to 100,000 meters. With that all figured out, when we
calculate this fraction, to three significant figures, we find a result of 3.34
newtons. That’s the magnitude of the
gravitational force experienced by object A due to the very large object.
Now, let’s consider the next
question in this exercise.
What is the acceleration of object
A toward the very large object? Give your answer to three
significant figures.
Okay, so, we’ve calculated the
gravitational force between these objects. And now, we want to calculate the
acceleration object A experiences. We can start doing this by
recalling Newton’s second law of motion. This law has it that the net force
on an object is equal to that object’s mass times its acceleration. Acceleration, in the case of object
A, is just what we want to solve for. And here’s how we can do it. As we saw, Newton’s second law says
that the net force on an object is equal to that object’s mass times its
acceleration.
When we think about the forces
acting on object A, we can see that there are two. One is the gravitational force due
to object B and another is the gravitational force due to the large spherical
object. But here’s the thing. The force between A and B is much,
much, much smaller than the force between A and the large spherical object. This is because the mass of our
large object absolutely dwarfs the mass of object B. We can say then that the
gravitational force on object A due to the very large object is effectively the only
force on object A.
And this means we can say that this
equation is equal to the mass of object A multiplied by the acceleration of object
A. We’ll call it 𝑎 sub A. And note that we got this
expression here from Newton’s second law. But now, as we consider this
equation, notice that the mass of object A appears on both sides, and therefore we
can cancel it out. If we do that, this is the equation
that results. The acceleration of object A is
equal to the universal gravitational constant times the mass of our very large
object divided by the distance between the centers of mass of object A and our very
large object squared.
Now, as we consider this part of
our expression here, we can see that it’s actually equal to this expression if we
drop out the mass of our object, in this case, object A. With that mass gone, see that we
have the gravitational constant, uppercase 𝐺, multiplied by the mass of our large
object divided by 𝑟 squared. Which are identical to the factors
we see here on the right-hand side of this equation. So then, if we calculate all this,
we’ll solve for the acceleration of object A due to the very large object. To three significant figures, it’s
0.667 meters per second squared. That’s the acceleration of object A
toward the very large object.
Now, let’s consider the next part
of our exercise.
What is the magnitude of the
gravitational force experienced by object B due to the very large object? Give your answer to three
significant figures.
Okay, whereas before, we were
considering the gravitational force between object A and the large spherical object,
now we’re considering that force between the large object and object B. Once again, we’ll use Newton’s law
of gravitation to solve for this force. But this time, instead of solving
for the force experienced by object A, we’ll solve for that experienced by object
B. And we’ll say that this object has
a mass 𝑚 sub B. Because both objects A and B are
the same distance away from the center of mass of our large spherical object, we
won’t change 𝑟, the distance between the centers of mass of the two masses in our
equation. Which means that when we go to
calculate this force, 𝐹 sub B, using an expression like this, all we’ll need to do
is substitute the mass of object B in where we used to have the mass of object
A.
We see that this mass is 100
kilograms. And so, now, we have an expression
which, when we solve it, will give us this force experienced by object B due to the
gravitational attraction from the large spherical object. To three significant figures, this
force is 66.7 newtons. Note that this is larger than the
force magnitude experienced by object A. In fact, it’s 20 times larger
because object B has a mass 20 times larger than object A.
Now that we found this out, let’s
consider the last part of our question.
What is the acceleration of object
B toward the very large object? Give your answer to three
significant figures.
To answer this question, once
again, we’ll use Newton’s second law of motion. And just like we did for object A,
we’ll also assume that the gravitational force between objects A and B is negligibly
small compared to that force between object B and the very large object. This means we can say that the mass
of object B multiplied by its acceleration is equal to the gravitational force
experienced by object B due to the very large object. So then, we can go to our equation
for that force, and we can equate it to the mass of object B times the acceleration
of object B. We’ll call it 𝑎 sub B.
And then, like we saw earlier, the
mass of our object, in this case, object B, is common to both sides. And that means that the
acceleration experienced by object B is equal to capital 𝐺 times the mass of our
very large object divided by the distance between B and our very large object
squared. And this expression here is equal
to this expression here if we ignore this mass of our object B. And so, to solve for 𝑎 sub B,
we’ll multiply the universal gravitational constant by the mass of our large object
divided by the distance between that object and B squared.
To three significant figures, this
is 0.667 meters per second squared. Note that this is the same as the
acceleration experienced by object A. And this comes back to the fact
that our equation for object acceleration does not depend on the mass of the object
whose acceleration we’re considering. So, this is how object B and object
A, and any other object, will accelerate toward the very large object.
