Video: Determining the Effect of Density and Radius on the Gravitational Acceleration Produced near a Planetary Surface

The acceleration due to gravity on the surface of a planet is three times as large as it is on the surface of Earth. The density of the planet is known to be twice that of the Earth. What is the radius of this planet in terms of Earth’s radius?

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Video Transcript

The acceleration due to gravity on the surface of a planet is three times as large as it is on the surface of Earth. The density of the planet is known to be twice that of the Earth. What is the radius of this planet in terms of Earth’s radius?

We are asked to solve here for the radius of this unknown planet, which we’ll call π‘Ÿ sub 𝑝. We can begin our solution by writing down the relationships that we’re given in the problem statement. First, we’re told that gravity on this unknown planet is three times as strong as the acceleration due to gravity on Earth. We’ll write that relationship as 𝑔 sub 𝑝, the acceleration due to gravity on the planet, is equal to three times 𝑔 sub 𝑒, the acceleration due to gravity on Earth. We’re are also told that the density of this planet, which we’ll refer to as 𝜌 sub 𝑝, is equal to two times the density of Earth, which we’ll call 𝜌 sub 𝑒.

Let’s begin by recalling the relationship that tells us the acceleration due to gravity on a given planet. If we have a spherical planet with mass 𝑀 and a radius π‘Ÿ, then the acceleration due to the gravity created by this planet on the surface of the planet, which we’ll call lowercase 𝑔, is equal to uppercase 𝐺, the universal gravitational constant, multiplied by 𝑀 divided by π‘Ÿ squared.

Any spherical mass of assumed uniform density follows the same relationship, which means we can use this relationship to help us solve for π‘Ÿ sub 𝑝, the radius of our unknown planet, in terms of π‘Ÿ sub 𝑒, the radius of the Earth. Let’s write out equations for 𝑔 sub 𝑝, the gravitational acceleration on the planet, and 𝑔 sub 𝑒, the gravitational acceleration on Earth. 𝑔 sub 𝑝, gravity on our unknown planet, is equal to capital 𝐺, the universal gravitational constant, multiplied by the mass of our unknown planet divided by its radius squared.

And we’re told in the problem statement that this is equal to three times the gravity on Earth. And we can substitute in for 𝑔 sub 𝑒 according to our equation for acceleration due to gravity. We now have an equation that has π‘Ÿ sub 𝑝 that we’re working to solve for and π‘Ÿ sub 𝑒, the radius of the Earth. The only issue is we have π‘š sub 𝑝, the mass of the planet, which is unknown. This is where the second relationship given in the problem statement becomes useful to us.

The density of the unknown planet is twice the density of Earth. Recall that density is defined. Knowing the density is defined as the mass of an object divided by its volume, we can use this equation to write out density equations for our unknown planet 𝑝 and the Earth. The density of the Earth is equal to the Earth’𝑠 mass divided by its volume and likewise for the unknown planet 𝑝.

As we look back at our equation for gravitational acceleration, what we’d like to do is replace π‘š sub 𝑝 and π‘š sub 𝑒, the mass of the unknown planet and the mass of the Earth, so that we can solve for this ratio purely in terms of the radii of the planets. We can move in that direction by using our density relationship to solve for the mass of the Earth and the mass of the planet in terms of their density and volume.

Let’s multiply both sides of the density of the Earth equation by the volume of the Earth. When we do that, the volume of the Earth term cancels out of the right side, leaving us with an equation that reads the mass of the Earth is equal to the volume of the Earth times its density. Likewise, when we multiply the density of the unknown planet equation, both sides by the volume of that unknown planet, 𝑣 sub 𝑝, then the term 𝑣 sub 𝑝 cancels out of the right side of our equation. And we’re left with an equation that reads the mass of the unknown planet equals the volume of the unknown planet multiplied by its density.

Now, let’s recall that the volumes we’re talking about are the volumes of two spheres. And we know the volume of a sphere in general. That is given as four-thirds πœ‹ times the radius of the sphere cubed. We can now use this fact to substitute in for 𝑣 sub 𝑒 and 𝑣 sub 𝑝, the volumes of the Earth and unknown planet, respectively, in terms of their radii.

We now have the mass of the Earth and the mass of our unknown planet in terms of their density and radii. Before substituting these values back into our equation for acceleration, we can make one more substitution. Recall that the density of our unknown planet is equal to twice the density of the Earth. That means we can replace 𝜌 sub 𝑝 with two times 𝜌 sub 𝑒.

Now, we’re ready to substitute in for π‘š sub 𝑝 in our equation for gravitational acceleration. With that substitution complete, we see there’re several terms we can cancel out from both sides of our equation. Capital 𝐺, the universal gravitational constant, cancels as does the factor four-thirds πœ‹. The density of the Earth 𝜌 sub 𝑒 also appears on both sides of the equation and therefore can be cancelled out.

And as we look at the radii on either side of our equation, we have also π‘Ÿ sub 𝑝 cubed divided by π‘Ÿ sub 𝑝 squared on the left side and on the right side, π‘Ÿ sub 𝑒 cubed divided by π‘Ÿ sub 𝑒 squared. In both these cases, the denominator will cancel out with two factors of the radii of the numerator, leaving a power of the radii in the numerator of one.

So when all the factors that are common are cancelled out or simplified, we’re left with an equation that reads two times π‘Ÿ sub 𝑝, the radius of the unknown planet, equals three times π‘Ÿ sub 𝑒. Dividing both sides by two, which cancels that term out of the left-hand side of our equation, we’re left with an expression that reads π‘Ÿ sub 𝑝 is equal to three-halves π‘Ÿ sub 𝑒. This is the value of the radius of the unknown planet in terms of the radius of the Earth.

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