Video: Calculating the Orbital Speed of a Satellite

For a satellite to follow a circular orbit around Earth at a radius of 10,000 km, what orbital speed must it have? Use a value of 5.97 × 10²⁴ kg for the mass of Earth and 6.67 × 10⁻¹¹ m³⋅kg⁻¹⋅s⁻² for the value of the universal gravitational constant. Give your answer to 3 significant figures.

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Video Transcript

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.

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