Video: The Gravitational Field Inside Planets

In this video, we will learn how to calculate the magnitude of the acceleration due to gravity at a point inside of a planet.

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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 weโ€™re 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 Newtonโ€™s universal law of gravitation that the force of gravitational attraction between two objects is equal to ๐บ, the universal gravitational constant, times ๐‘š one, the mass of the planet, times ๐‘š two, the mass of the object, divided by ๐‘Ÿ squared, where ๐‘Ÿ is the distance between the center of the planet and the center of the object.

The second equation we need to use is Newtonโ€™s second law, where the net force experienced by an object is equal to the mass of the object times the objectโ€™s 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‘š two as thatโ€™s what we called the objectโ€™s 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‘š two ๐‘Ž in our equation. To isolate the acceleration, we have to multiply each side by one over ๐‘š two. This will cancel out the ๐‘š two on the left side of the equation as well as on the right side of the equation, leaving us with the equation ๐‘Ž, the acceleration due to gravity at any point outside the planet, is equal to big ๐บ, the universal gravitational constant, times ๐‘š one, the mass of the planet, divided by ๐‘Ÿ squared, where ๐‘Ÿ 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 weโ€™re 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‘Ÿ 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? Letโ€™s compare what we just went over to what would happen inside the planet.

๐‘Ž, the acceleration due to gravity inside the planet, is equal to four-thirds times ๐บ, the universal gravitational constant, times ๐œŒ, the density of the planet, times ๐œ‹ times ๐‘Ÿ, 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‘Ÿ cannot be bigger than the distance between the center of the planet and its surface, as represented by big ๐‘… in our diagram. Here, weโ€™ll only consider cases where little ๐‘Ÿ is less than big ๐‘…. When little ๐‘Ÿ equals big ๐‘…, our inside-of-planet and outside-of-planet equations for acceleration become the same. When little ๐‘Ÿ is bigger than big ๐‘…, our equation no longer holds true and we must use the acceleration equation that we saw previously.

Letโ€™s apply this formula to an example problem. Letโ€™s 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‘Ž equals four-thirds big ๐บ๐œŒ๐œ‹๐‘Ÿ.

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œŒ, we use 5.52 times 10 to the third kilograms per meters cubed. For ๐œ‹, we use 3.14. As all of our other variables have been given to us to three significant figures, we round ๐œ‹ to three significant figures. And for our distance ๐‘Ÿ, 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? Letโ€™s 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‘Ÿ 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‘Ž over ๐‘Ÿ by dividing both sides by ๐‘Ÿ, we would get the slope of this diagonal line to be four-thirds ๐บ๐œŒ๐œ‹, where four-thirds, ๐บ, and ๐œ‹ 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.

Letโ€™s 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 theyโ€™re 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, weโ€™d 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 letโ€™s go through a few examples that use our formula and analyze graphs.

What is the acceleration due to gravity at a distance from Earthโ€™s center of mass that is equal to half the Earthโ€™s 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 weโ€™re within Earthโ€™s 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‘Ž, the acceleration due to gravity, equals four-thirds ๐บ, the universal gravitational constant, ๐œŒ, the density of the planet, ๐œ‹๐‘Ÿ, where ๐‘Ÿ is the distance from the center of the planet to the position specified. Four-thirds, ๐บ, and ๐œ‹ are all constants, so that leaves ๐œŒ and ๐‘Ÿ as variables we need to define from our problem.

For ๐บ, we use 6.67 times 10 to the negative 11th newtons times meter squared per kilogram squared. ๐œŒ is given to us in our problem as 5510 kilograms per meters cubed. For ๐œ‹, weโ€™ll 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, weโ€™re told that itโ€™s half of Earthโ€™s radius, and weโ€™re 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 Earthโ€™s center of mass that is equal to half the Earthโ€™s 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, weโ€™ve 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 ๐บ, the universal gravitational constant, ๐œŒ, the density of the planet, ๐œ‹๐‘Ÿ, where ๐‘Ÿ is the distance from the center of the planet to a point within the surface. The acceleration outside of a planet is equal to ๐บ, the universal gravitational constant, times ๐‘š, the mass of the planet, divided by ๐‘Ÿ squared, where ๐‘Ÿ is the distance from the center of a planet to a point thatโ€™s outside the surface.

In our first question, weโ€™re 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, itโ€™s 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, letโ€™s look at the equation for volume. The volume of a sphere, as weโ€™ve assumed each planet is a sphere, is equal to four-thirds ๐œ‹๐‘Ÿ cubed. Four-thirds and ๐œ‹ 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, letโ€™s look at the field outside of the planet. For any radial distance, letโ€™s 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 letโ€™s check to make sure thatโ€™s correct.

The slope of the line inside of our planet, ๐‘Ž divided by ๐‘Ÿ, is equal to four-thirds ๐บ๐œŒ๐œ‹, where four-thirds, ๐บ, and ๐œ‹ 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, letโ€™s 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‘Ž equals four-thirds ๐บ๐œŒ๐œ‹๐‘Ÿ 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 planetโ€™s center to determine dimensions of the planet. Interpret 3D graphs of acceleration due to gravity against position.

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