Suppose a satellite is in a circular orbit not far above Earth...

Question

Suppose a satellite is in a circular orbit not far above Earth’s surface, and circles the Earth about once every 92 minutes. Find the centripetal acceleration of the satellite in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth’s surface.

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve for this problem. What is the centripetal acceleration of a space station? It orbits earth in a steady circular path just above the atmosphere. If it completes its orbit every 95 minutes express the result in terms of G the gravitational acceleration at earth's surface. So that's our angle. Our angle is, we're trying to figure out what the centripetal acceleration of this particular space station is as it orbits earth. And that's our final answer that we're ultimately trying to solve for is what is the centripetal acceleration? Awesome. We're also given a figure from the problem itself. So the problem itself gives us a figure to look at. So as we can see, we have a black circle going all the way around. And then in our upper right hand corner, we have our green space station that's orbiting around earth and earth is the second circle inside of our black circle. And as you can see, it's a globe and it has blue for the ocean and then black for the continents. And we can see from the center of earth going out to where the space station is orbiting. We have our orbital radius of 6680 kilometers. And then we also have our radius of earth which is capital R subscript E which is 6380 kilometers. And then we have our distance H which is from earth out to our orbital C where the, where our space station is orbiting. And that is our distance H Awesome as shown in our figure itself. And then we're told in red that we have an orbit of 95 minutes. Awesome. And that's how long it takes to complete one full orbit around earth for our space station. OK. So now that we have a visualization of what's happening in this particular problem, let's look at our multiple choice sensors to see what our final answer might be. And let us note that they're all in terms of G. So A is 0.20 B is 0.33 C is 0.40 and D is 0.83. OK. So first off, let us recall and use the equation to solve for the centripetal acceleration. And let us call this equation one. So the centripetal acceleration A is equal to V squared divided by R where V is the speed of the space station and lowercase R is our orbital radius. So let us note that we will use and let's write this as a side note in blue because this is an important side note, we will use that R is approximately equal to capital re which once again, capital R subscript capital E is our orbital radius of earth. So we're saying that the radius of our orbit of the space station is approximately equal to the radius of earth or OK. So with the radius of earth, so moving right along here and once again, re is our radius of earth. So now we must note that we are not given the speed of the space station. So we must solve for the speed of the space station in the terms of its period. Therefore, we can recall and use the following equation. And let's call this equation two, the equation two states that V is equal to two multiplied by pi multiplied by R divided by capital T where capital T is the period. So now, at this point, we need to set equations one and two equal to each other. And we can say that A is equal to two multiplied by pi multiplied by R squared. We're actually I said set it equal to each other. But what I meant to say is is that we're taking equation A and we're plugging in our value for V into A. So that's what we're doing to be specific OK. So two multiplied by pi multiplied by R all to the power of two because we're plugging in, once again, we're plugging our value of V back into equation one. And now we need to divide by R multiplied by T squared. And when we simplify, we can say that this is all equal to four multiplied by pi squared, multiplied by R all divided by T squared. OK. So now moving right along, we must recall and use the equation to help us solver the centripetal acceleration of the space station. And let's call this equation three. So we'll call the centripetal acceleration AC and AC is equal to four multiplied by pi squared multiplied by R all divided by T squared. And once again, capital T is our orbital period. Let us note that the orbit radius will be the sum of earth's radius, which is as we should recall, 6380 kilometers plus the 300 kilometer orbital height or the orbit height is there's a 300 kilometer orbit height. So let us note that and let's write this in blue since this is another side note. So let's take a little side note again, up top here in blue that are, is equal to 6380 kilometers plus 300 kilometers is all equal to. So 6380 kilometers plus 300 kilometers will give us 6680 kilometers. And let's also like write a little note here that T our period is equal to 95 minutes. Awesome. So now at this point, we need to plug in these values into equation three to help us solve for AC. But let us first quickly convert our period from minutes to seconds to get it in the proper units. So let us note that the period T is equal to 95 minutes. And using dimensional analysis, we should be able to recall that there's 60 seconds in one minute. So when we plug that into our calculator, we will find that 95 minutes is equal to 5700 seconds. So now we can officially solve for AC. So AC is equal to four multiplied by pi squared multiplied by. And let's note that we have to convert kilometers to meters. So when we plug in our value for R, we will use 6.68. So like we did for our period, you can use dimensional analysis to find this answer. I'm just going ahead and giving you the conversion. So it's 6.68 multiplied by 10 to the power is 6 m. And once again, we took our radius value which was initially in kilometers and we converted it to meters. And now we need to divide everything by our period squared, which is 5700 seconds squared. So when we plug that into our calculator, and we round to two decimal places, we will get 8.12 and its units will be meters per second squared. Awesome. So now at this point, we need to convert our meters per second squared to G units. So we need to take 8.12 m per second squared. And we need to use dimensional analysis again and note that in one G there is 9.81 m per second because that is the acceleration due to gravity on earth. So when we plug that into our calculator and we round to two decimal places, we will get 0.8 three GS and that's it we solve for this answer or, or found our final answer. Hooray, we did it. So looking at our multiple choice answers, the correct answer has to be the letter D 0.83 G and that's it we've solved for this problem. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Bye.

Suppose a satellite is in a circular orbit not far above Earth’s surface, and circles the Earth about once every 92 minutes. Find the centripetal acceleration of the satellite in its orbit. Express your answer in terms of g, the gravitational acceleration at the Earth’s surface.

Key Concepts

Centripetal Acceleration

Gravitational Acceleration (g)

Orbital Period

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