### Video Transcript

In this video, we will learn how to
calculate the magnitude of the acceleration due to gravity at a point inside of a
planet. For this video, we will only be
looking at spherical planets with uniform density to make our calculation
simpler. Before we look at the acceleration
due to gravity inside a planet, we must first refresh our memory about acceleration
due to gravity outside a planet.

When weโre looking for the
gravitational field strength experienced by an object at a specific distance from a
planet, we need to use two equations. The first is Newtonโs universal law
of gravitation that the force of gravitational attraction between two objects is
equal to ๐บ, the universal gravitational constant, times ๐ one, the mass of the
planet, times ๐ two, the mass of the object, divided by ๐ squared, where ๐ is the
distance between the center of the planet and the center of the object.

The second equation we need to use
is Newtonโs second law, where the net force experienced by an object is equal to the
mass of the object times the objectโs acceleration. If we are to apply the second
equation to our object, then we would say the net force is the force of
gravitational attraction between the planet and the object. And the mass will be ๐ two as
thatโs what we called the objectโs mass earlier.

Since we said that the force of
gravitational attraction is the same as the net force in this case, we can replace
๐
g with ๐ two ๐ in our equation. To isolate the acceleration, we
have to multiply each side by one over ๐ two. This will cancel out the ๐ two on
the left side of the equation as well as on the right side of the equation, leaving
us with the equation ๐, the acceleration due to gravity at any point outside the
planet, is equal to big ๐บ, the universal gravitational constant, times ๐ one, the
mass of the planet, divided by ๐ squared, where ๐ is the distance between the
planet and the position we are at.

This equation shows that the
acceleration due to gravity outside of the spherical planet is directly proportional
to the mass of the planet and that the acceleration has an inverse square
relationship to the distance between the center of the planet and the position in
which weโre taking our measurement. If we were to plot a graph of
acceleration due to gravity versus distance from the center of the planet, it would
look like this, with ๐ being the distance from the center of our planet to the
surface of the planet. But what does the acceleration due
to gravity look like between the center of the planet and the surface? Letโs compare what we just went
over to what would happen inside the planet.

๐, the acceleration due to gravity
inside the planet, is equal to four-thirds times ๐บ, the universal gravitational
constant, times ๐, the density of the planet, times ๐ times ๐, the distance
between the center of the planet and the position inside the surface of the
planet. In this video, we will only be
dealing with densities of the planet that are uniform. It is important to note that little
๐ cannot be bigger than the distance between the center of the planet and its
surface, as represented by big ๐
in our diagram. Here, weโll only consider cases
where little ๐ is less than big ๐
. When little ๐ equals big ๐
, our
inside-of-planet and outside-of-planet equations for acceleration become the
same. When little ๐ is bigger than big
๐
, our equation no longer holds true and we must use the acceleration equation that
we saw previously.

Letโs apply this formula to an
example problem. Letโs say that we have a spherical
planet with a uniform density that matches the average density of Earth, 5.52 times
10 to the third kilograms per meters cubed. And we wanna know what the
acceleration due to gravity is at a distance of 4.32 times 10 to the fourth meters
from the center of the planet, assuming that the radius of the planet is greater
than this distance. To solve for the acceleration, we
use the formula ๐ equals four-thirds big ๐บ๐๐๐.

Plugging in our variables, we use
6.67 times 10 to the negative 11th newtons meter squared per kilogram squared for
the universal gravitational constant. For ๐, we use 5.52 times 10 to the
third kilograms per meters cubed. For ๐, we use 3.14. As all of our other variables have
been given to us to three significant figures, we round ๐ to three significant
figures. And for our distance ๐, we use
4.32 times 10 to the fourth meters. When we multiply out our values, we
get an acceleration due to gravity at this position of 6.66 times 10 to the negative
second meters per second squared.

Remember that graph we drew
earlier? Letโs add in the acceleration due
to gravity when we are anywhere inside the planet. We can see from our equation that
the distance between the center of the planet and our position is directly
proportional to the acceleration due to gravity. On the graph, this will yield the
diagonal line from the origin to the accompanying point for the acceleration due to
gravity at the surface of the planet. From the graph, we can see that the
acceleration due to gravity at the center of the planet, as represented by the
origin, is zero, which should make sense because if we plug in zero for ๐ in our
equation, anything multiplied by zero is still zero. So our acceleration due to gravity
would be zero.

Looking back at our formula, we can
see that if we wanna find the variable form of the slope which is rise,
acceleration, divided by run, distance, or ๐ over ๐ by dividing both sides by ๐,
we would get the slope of this diagonal line to be four-thirds ๐บ๐๐, where
four-thirds, ๐บ, and ๐ are all constants. This means if we were given
numbers, we could determine the density of our planet by finding the slope from our
graph. Qualitatively, we can compare the
densities of different planets if we graph their accelerations due to gravity. What would the graph look like for
a planet that has a larger density but the same radius as the planet that we chose
to represent?

Since density is proportional to
the slope, it would be a steeper slope as represented by the yellow line. At the same distance big ๐
, the
new planet transitions into the inverse square relationship just as the pink planet
did. What about if the planet had a
smaller density when compared to our sample planet, but still the same radius? Once again, the density of the
planet is proportional to the slope, so a smaller density would have a smaller
slope, or more shallow as represented by the blue line. Once again, at the distance big ๐
,
our graph transitions into the inverse square relationship curve, just as it did for
our sample planet in pink and our previous planet with the larger density.

Letโs take it one step further and
analyze the acceleration due to gravity using a 3D model. The tip of the cone labeled D in
this model is in the center of the planet and represents an acceleration due to
gravity of zero. The cone bottoms out where the
surface of the planet is, labeled B in this model, and has a maximum value for the
acceleration due to gravity as we saw in the previous graph. The 3D model has acceleration due
to gravity becoming closer to zero as the distance from the surface of the planet
increases. Although the points outside our
point D on the graph are lower, so one might think they have a negative
acceleration. In fact, the vertical axis has
positive values below zero, so theyโre actually positive. So the value of the gravitational
field will be smaller as we go from position E to C to A.

If we were to rank the acceleration
from greatest to smallest, weโd have B at the surface of the planet, then E a little
ways away from the planet, C farther from the planet, A even farther from the
planet, and D at the center of the planet. This model provides a nice visual
representation of what is happening with the acceleration due to gravity. Now letโs go through a few examples
that use our formula and analyze graphs.

What is the acceleration due to
gravity at a distance from Earthโs center of mass that is equal to half the Earthโs
radius? Assume that Earth is a perfect
sphere with a radius of 6370 kilometers and a constant density of 5510 kilograms per
meters cubed. Give your answer to three
significant figures.

In the problem, we are told that
weโre within Earthโs surface. Therefore, we must use an equation
where we find acceleration due to gravity inside a planet. The equation to find the
acceleration due to gravity inside a planet is ๐, the acceleration due to gravity,
equals four-thirds ๐บ, the universal gravitational constant, ๐, the density of the
planet, ๐๐, where ๐ is the distance from the center of the planet to the position
specified. Four-thirds, ๐บ, and ๐ are all
constants, so that leaves ๐ and ๐ as variables we need to define from our
problem.

For ๐บ, we use 6.67 times 10 to the
negative 11th newtons times meter squared per kilogram squared. ๐ is given to us in our problem as
5510 kilograms per meters cubed. For ๐, weโll take it out to three
significant figures as 3.14 as our problem asks us to give our answer to three
significant figures. For the distance from the center,
weโre told that itโs half of Earthโs radius, and weโre told that the radius of Earth
is 6370 kilometers. When we multiply one-half by 6370
kilometers, we get 3185 kilometers. Before we multiply out our values,
we need to first check our units. Our distance is given in
kilometers, but our density has meters in it, and so does our universal
gravitational constant. Therefore, we need to convert from
kilometers to meters.

We need to recall that kilo- is a
prefix that means 1000 or 10 to the third. Therefore, we can replace 3185
kilometers with 3185 times 10 to the third meters. Now that all of our units are in
agreement, we can multiply out our values to get our final answer. When we multiply out our values and
round to three significant figures, we get an acceleration due to gravity of 4.90
meters per second squared. The acceleration due to gravity at
a distance from Earthโs center of mass that is equal to half the Earthโs radius is
4.90 meters per second squared.

The graph shows how the
acceleration due to gravity varies with radial distance inside and outside of three
planets. Each planet is a perfect sphere and
has a constant density. Which of the following is true
about all of the planets? (A) All of the planets have the
same surface gravity. (B) All of the planets have the
same radius. (C) All of the planets have the
same volume. (D) All of the planets have the
same mass. (E) All of the planets have the
same density. Which planet has the largest
radius? (A) They all have the same radius,
(B) planet B, (C) planet A, (D) planet C.

On our graph, our planets are
represented by three different colors. Planet A is red, planet B is blue,
and planet C is green. All of the planets overlap on the
diagonal line that runs from the origin to the radial distance of 8000
kilometers. We can see that there are two
distinct shapes on the graph. The diagonal line is the
relationship between the acceleration due to gravity and radial distance inside of
the planets. On the graph, weโve labeled it to
start and end for planet C. The planets end at three different radial distances:
6000 kilometers for planet A, 7000 kilometers for planet B, and 8000 kilometers for
planet C.

The curved portion of the graph
represents outside of the planet and has once again been labeled on the graph for
planet C. We should know that the shape of the curve for each planet outside is
similar. The shapes of the graph come from
the equations for the acceleration due to gravity inside of a planet and the
acceleration due to gravity outside of a planet. The acceleration inside the planet
is equal to four-thirds ๐บ, the universal gravitational constant, ๐, the density of
the planet, ๐๐, where ๐ is the distance from the center of the planet to a point
within the surface. The acceleration outside of a
planet is equal to ๐บ, the universal gravitational constant, times ๐, the mass of
the planet, divided by ๐ squared, where ๐ is the distance from the center of a
planet to a point thatโs outside the surface.

In our first question, weโre asked
to determine what is true for all of the planets. If all the planets have the same
surface gravity, then we would expect them to have the same acceleration due to
gravity at the surface. The surface on our graph is the
point at which our planet switches from inside to outside. That would be in the three
different locations represented by the vertical dotted lines. For planet A, the surface is at
6000 kilometers, planet B is at 7000 kilometers, and planet C 8000 kilometers. The acceleration due to gravity at
the surface for planet A is just above nine meters per second squared. At planet B, itโs just below 11
meters per second squared, and planet C is just above 12 meters per second
squared.

Since all three planets have
different accelerations due to gravity at their surface, we can say that answer
choice (A) is not true. If all the planets have the same
radius, then on the graph, they would switch from inside to outside at the same
radial distance. However, as we discussed earlier, A
is at 6000, B is at 7000, and C is at 8000 kilometers. Therefore, answer choice (B) is
also not true.

To determine whether or not they
have the same volume, letโs look at the equation for volume. The volume of a sphere, as weโve
assumed each planet is a sphere, is equal to four-thirds ๐๐ cubed. Four-thirds and ๐ are both
constants. That means that, depending on the
radius, our volume may be the same or different for each planet. As we just stated, the three
planets have different radii. Therefore, they have different
volumes. So answer choice (C) is also not
true.

To determine if the planets have
the same mass, letโs look at the field outside of the planet. For any radial distance, letโs
choose 12000 kilometers from the center of each planet, we can see that each of them
will have a different acceleration due to gravity. In our formula outside of the
planet, we can see that the acceleration due to gravity is based on the mass, along
with the distance away. Since each planet has a different
acceleration due to gravity at the same radial distance, they must each have
different masses, with planet C being the biggest mass and planet A being the
smallest mass. We can therefore eliminate answer
choice (D), leaving us with answer choice (E) that they all have the same
density. But letโs check to make sure thatโs
correct.

The slope of the line inside of our
planet, ๐ divided by ๐, is equal to four-thirds ๐บ๐๐, where four-thirds, ๐บ, and
๐ are all constants. Therefore, if our planets have the
same densities, then they all have the same slope. Looking at our graph, we can see
that all three planets do indeed have the same slope. Answer choice (E), all of the
planets have the same density, is true.

For the second question, which
planet has the largest radius, letโs go back to our graph. Remember that we said that the
radius is going to be the radial distance at which we go from inside our planet to
outside our planet. And we said that this happened at
6000 kilometers for planet A, 7000 kilometers for planet B, and 8000 kilometers for
planet C. 8000 kilometers is the largest number, which belongs to planet C. The
planet with the largest radius is answer choice (D) planet C.

Key Points

The acceleration due to gravity
inside a planet of constant density is proportional to the distance away from the
center of the planet. The acceleration due to gravity at
the center of a planet is zero. Use the formula ๐ equals
four-thirds ๐บ๐๐๐ to find the acceleration due to gravity at a point within a
planet of constant density. Interpret graphs of acceleration
due to gravity against radial distance from a planetโs center to determine
dimensions of the planet. Interpret 3D graphs of acceleration
due to gravity against position.