Satellite Motion: Speed & Period: Study with Video Lessons, Practice Problems & Examples

Satellites maintain circular orbits due to a balance between gravitational force and tangential velocity. The orbital speed, GMr, is crucial for this motion. The orbital period, T^2 = (4 × π^2 × r^3) / GM, relates time to distance. As distance increases, velocity decreases, while the period increases, reflecting the weaker gravitational pull at greater distances.

Satellite Speed

Hey, guys. So remember when we were talking about satellite motion, we said that there was a specific velocity a satellite needed in order to go in a perfectly circular orbit. Well, you're going to need to know how to use that formula and calculate that velocity. So let's go ahead and cover that in this video. We have a satellite that's out at some distance away from the Earth. That center of mass distance is little r. At that distance, there is a gravitational force that constantly tries to pull it back towards the Earth. So the question is why does it just come crashing into the surface? Remember that the satellite is actually falling towards the Earth, but it has a tangential velocity that basically keeps it going in a circle. This thing is constantly falling around the Earth, and the Earth is trying to pull it backward, but that tangential velocity keeps it going, so the Earth is constantly curving beneath it. Okay. And so, for a satellite in circular orbit, that gravitational force is actually what's causing it to go in a circle. That gravitational force keeps the satellite going in uniform circular motion. And the relationship between that speed, which is the orbital speed, and the distance, which is that little r, is the v_sat equation. That is the square root of capital G capital M over little r. I just want to remind you guys that that capital M is actually the mass of the big planet that it's going around, not the mass of the satellite. And that little r is not the radius, it's the orbital distance. It's that little r distance away. Sometimes you're going to need to know where that equation comes from, so I can actually go ahead and work it out for you really quickly. How do we get the velocity from uniform circular motion? Well, remember that if this is the force that's acting on it and it's going uniform circular motion, then it has a centripetal acceleration, so we can start from F = ma to get to this. We know that F = ma, but all these forces are centripetal. So the sum of all centripetal forces equals m a_c. Now we know the only force that's acting on this is the force of gravity, and that is m. The a_c comes v² over r. So this is where this velocity actually comes from. Now we know what this force of gravity is. It's just G big M little m over r², that's just Newton's Law of Gravity. And that's equal to mv² over r. So to figure out what this v² is, let's go ahead and simplify this equation. I've got little m that appears on both sides, so I can cancel that out, and I've got little r that also appears on both sides. And so what I'm left with is I'm left with capital G capital M over r equals v². So, if I take the square roots, I'm just going to get to that Gm over r. I'm going to get to that v_sat equation. So again, that is, the relationship between the orbital speed and the orbital distance v_sat in order to keep this thing going in circular motion. So for instance, this satellite out here that's at some distance r has an exact v in order to keep it going in a perfect circle. Anything more or less than that, it's not going to be perfectly circular. And also if I wanted to change this orbit, if I wanted to go out farther or if I wanted to push this thing in closer, my v_sat would have to change in order to keep this thing traveling in a perfect circle. So that's what that means. Alright, guys. Let's go ahead and work out an example for this with the International Space Station. So we're asked to find the height of the International Space Station, which travels around the Earth. And we're told that the orbital speed is 7,670 meters per second in a nearly circular orbit. So anytime you see this word nearly circular, you're just going to assume that they're talking about a circular orbit, so you can use all these equations to do that. Okay. So what are we told? We're actually trying to figure out what the height of this thing is. But remember that whenever we're trying to find little h or big R, we're always going to find little r first. And then we can relate it using the r equals big R plus h formula. So, first, we have to find little r. So how am I going to do that? Which equation am I going to use? I'm trying to find what little r is, and I'm only told what the velocity of the satellite is. So I can use the v_sat equation to relate those two things because those are the only two variables that pop up in that equation. So let's start from the v_sat equation. So I've got v_sat equals square root of GM over r. So now, I want to actually get to what this r is, so then I can basically get to what h is. So I just have to, I'll go ahead and isolate that. I've got this r that's trapped in the denominator here, so I can lift the square roots by taking the square of both sides. So I've got v_sat² equals GM over r. And now if I want to get r by itself, I basically want this thing to come up, and I want the v_sat to come down. So these things are just going to trade places. So I've got that r equals GM over v_sat². And now I just have to make sure I have all of those numbers. Right? I have G, I have M, and I have v_sat². And that capital M, because I'm going around the Earth, is just the mass of the Earth, which I have in this table right here. So, plugging all that stuff in, I get 6.67 times 10 to the minus 11. And I've got the mass of the Earth, 5.97 times 10 to the 24th. Oops. Times 10 to the 24th, and then divided by 7,670. Just make sure that you square that in the denominator. And you should get 6.77 times 10 to the 6th, and that's in meters. But just remember that we're not quite done yet because that was, we've only solved, oops. We've only solved for little r. Now we have to go and plug it back into this equation to solve for h, which is the height. So we have little r equals big R plus h. So, if we want to find h, we have to set, we have to isolate h. So h is equal to little r minus big R. So I've got 6.77 times 10 to the 6th minus, what's this what's this big R? Well, this big R, if we're talking about the Earth, is just the radius of the Earth. So we're going to have them subtract that, 6.37 times 10 to the 6th, and we should get 4 times 10 to the 5th, which is about 400 kilometers. You can actually go ahead and Google this. If you Google the height of the International Space Station orbit, you'll find that it is about 400 kilometers on average. It goes up and down a little bit, but that's pretty much what it is on average, which is pretty cool. So we can use some F = ma and some simple equations to figure out how high the International Space Station actually orbits around the Earth, which is pretty cool. Alright, guys. Let me know if you have any questions with this.

Suppose that you used some geometry and kinematics to estimate that the Earth goes around the Sun with an orbital speed of approximately 30,000 m/s (60,000 mph), and that the Sun is approximately 150 million kilometers away from the Earth. Use this information to estimate the mass of the Sun.

• A. 1.42×10¹⁵kg
• B. 2.02×10²⁹kg
• C. 2.02×10³⁰kg
• D. 1.42×10¹⁷kg

Find speed of second satellite, given speed of first

Hey, guys. Let's check out this problem here. We're told 2 satellites are orbiting this planet and we're given a whole bunch of numbers. So let's go ahead and just start drawing a diagram. My diagram here is going to involve some planet and I have 2 satellites in circular orbit. So let me go ahead and draw my circular orbits. I've got one right there, and I've got another one that's going to be right there, let's say. Cool. So now I've got a mass. I've got a satellite that's over here. I'm told the mass of that satellite is equal to 68. And I'm told that the orbital radius, the orbital radius is r1=6×108. Remember that is that little r distance and that is the distance between the centers of mass. That's not big r and that's not h. This is r. But because it's the first one, because I have 2 satellites, I'm going to call this r1. And then I'm told that the velocity, the orbital velocity, which is that tangential velocity v1, is equal to 3,000. Right? Cool. So now I have another satellite that is some other distance away. So I'm going to call this m2, and I know the mass is 84, and then I've got the orbital radius of that thing. So then I've got here, the orbital radius is equal to, r2, and I know what that number is. Right? That's 9×108. Now I'm asked to figure out what is the orbital velocity of the 2nd satellite. So really, v2 is my target variable. So let's start there. So I've got v2 as my target variable. So how do I get that? Remember from our equations, we're going to use the vsat equation if we're looking for orbital velocity. So I've got v2, it's just going to be the square root of g×M and because I'm looking for v2, I'm going to be plugging in r2 into this. So let's take a look. This is my target variable right here. Let's take a look if I have everything else in the problem. G is just a constant. Then I've got the mass of the planets, but I don't have that variable. So I've got the radius, I've got the orbital radius, but I don't have big M. So how do we go about solving for that? Because I'm going to need that. I'm never told it in the question. So let's go over here and try to solve for big M. Let me go ahead and scroll down. So I've got big M equals something. How do we get big M? Well, we can use either forces. Let me look at our table of equations right here. We can either use forces or we can use accelerations or we can use satellite motion. But I'm not told anything about forces in the problem. I don't have any forces between 2 objects and I don't have any accelerations here. I know that it's circular motion and I know that I'm working with satellites, so let's use the other vsat equation. So remember we have 2 satellites here. So let's look at the other satellite, the first one. v1=g×M÷r1. Now the r1 of that other satellite. So if I take a look at this, I have the v1. I have g as just a constant, and I have r1. So I can actually go ahead and figure out what this big M is. So I'm going to go ahead and rearrange. I've got this thing in the square root, so I have to get rid of that square root by squaring both sides. So I've got v12=g×M÷r1. Now I'm just going to move everything over and solve for M. So I've got v12×r1÷g=M. So actually I have everything there. That orbital velocity for that first one is 3,000. Then I've got to square that. Then the orbital radius of the first satellite, we're told, is 6×108 and then we are just dividing by the gravitational constant. And if you do that, you should get 8.1×1025 and that's in kilograms. So now that we have that mass right here, we can actually plug it back into this equation for and solve for v2. So let's go ahead and do that. I've got v2=6.67×10-11×8.1×1025÷9×108, and that's in meters. So you go ahead and do that, you get the velocity is equal to 2,450 meters per second. So we've got that number right here. Notice that this number is actually lower than the orbital velocity of the first satellite, and that's because it's a farther distance away. So we can see from the vsat equation that if r, the orbital distance, is farther, then your velocity is going to be lower. So that actually just makes some sense. Let me know if you guys have any questions with that.

You throw a baseball horizontally while on the surface of a small, spherical asteroid of mass 7×10¹⁶ kg and diameter of 22km. What is the minimum speed so that it just barely goes around the asteroid without hitting anything?

• A. 8.9 m/s
• B. 14.6 m/s
• C. 115.3 m/s
• D. 20.6 m/s

Satellite Period

Hey, guys. We saw that the orbital velocity of a satellite was how fast the satellite travels in its orbit. There's another variable you need to know to solve problems, and that's the orbital period, which is the time that it takes to complete one orbit. So we saw that the satellite has a tangential velocity as it's going around in its orbit, but it takes some amount of time to actually complete one full orbit. That's called t. We actually relate that velocity to the period just by using circular motion and basic kinematics. How do we get t? How do we get time from velocity? We know from kinematics that v=distancetime. What's the distance that this thing completes in one orbit? Well, it's traveling in a circle of radius r, so that's just the circumference of that circle, 2πr. And if you do that in one full circle, then this t just becomes a capital T. This equation is really useful for us in satellite motion, so I'm actually going to write it here. −2πrT. Now we can actually solve for this T, this capital T right here. Most of the time you'll see it in its T squared form. This is 4πr3∕gm. You can actually get to it pretty quickly by using this equation, so let's go ahead and do it real quickly. V_sat=2πrT. So what happens is if you want T, you can get T=2πr∕V_sat. So we have this V_sat, but we already have another equation that will tell us. We can actually stick that in here, and we're going to get 2πr∕gm∕r. So this nasty formula with square roots and fractions; what'll happen is if you square it, it'll become a lot cleaner: 2 becomes the 4, π becomes π², r becomes r². And what happens to the square root is that when you square it, the whole thing just goes away, or the square root goes away. So you get gm over the little r. So we've got this situation where you have a fraction on a fraction, and what happens is that this denominator of the bottom fraction will actually go up and merge with that top little r. So what'll happen is 4πr3∕gm. So you get that equation. This equation actually has a name. It is called Kepler's third law. Kepler was a guy who was studying the motions of planets in our solar system, and he noticed a relationship between the orbital period of all the planets and the distances from the sun. It's pretty cool. So we've got these three equations and these are all of our satellite motion equations. So what happens if the distance increases? Alright. So what happens if r increases? Well, we can use, how does v change? So we have these 2 V_sat equations, and if the little r is changing, how does this equation change? It's a little bit tricky to tell just by using this one because you have all 3 variables present in these, and we don't know how T changes yet. So instead, let's use the other V_sat equation. So as the r increases in the denominator, the V_sat has to decrease. Alright. And again, let's not use this equation for T because we have a relationship just between T and r. So it's going to be easier to relate those 2 by this equation. Now r is in the numerator here. So if r goes up, then T has to increase as well. So that means that as r increases, your velocity decreases, but your period increases. This should make some sense because as you're going farther away, the force of gravity gets weaker on you, which means you don't have to go as fast to stay in a circle. On the other hand, if you were to go really close to the earth, the earth would be pulling on you really, really hard, but you have to go really fast not to crash into the earth. Because you're traveling in a larger circle as r increases, the time that it takes also should be increasing. And that's basically it. So let's go ahead and take a look at an example and use all of these equations together. So we've got the orbital period and speed of the International Space Station. So first thing is we're going to be calculating the orbital period. So let's take a look at our equations. So T equals w...

A satellite orbits at an orbital period of 2 hours around the Moon. What is the satellite's orbital altitude?

• A. 1.9×10⁶m
• B. 1.2×10⁵m
• C. 1.7×10⁶m
• D. 3.2×10⁵m

Find mass of planet, given moon speed and period

Hey, guys. Let's see if we can work up this problem together. So we have a moon in orbit around a planet. Right? So let's go ahead and draw a diagram. We have a planet like this and a moon that is orbiting around in a circular orbit. Just pretend that's perfectly circular. Right? And this moon has a speed around its planet, and we're told that the velocity is equal to 75,100. And we're also told that it takes 28 hours to go all the way around. So, all the way one rotation, we know that \(t = 28\) and that is hours. So, using that information, how can we find out what the mass of this planet is? We're finding the mass of the planet. That's the thing in the center. So, we're really looking for capital \( M \). So, how do we go about doing that? Well, let's see. We've got a whole bunch of equations involving forces and gravitational acceleration, but we don't have any information about forces or gravitational acceleration. But we do have the motion of a satellite, so we're going to use all these equations over here. So let's see. I've got orbital velocity and I also have orbital period. So really I can use any one of these equations to start off. I'm going to use the \( v_{\text{sat}} \) equation. So we've got \( v_{\text{sat}} = \sqrt{\frac{g \times M}{r}} \). And let's see. I know what the velocity of the satellite is, and I'm looking for big \( M \), which is my target variable. Gravitational constants is the number. So, if I can figure out what this little \( r \) distance is, I could find that. But I actually don't have any information about the orbital distance or height or anything like that. So let's see. Maybe this isn't the approach. Let's see if I can start off with my \( t_{\text{sat}}^2 \) equation, with the \( t^2 \) one. So let's start out with the \( t_{\text{sat}}^2 \) equation. Let's see if we have better luck there. So you've got \(\frac{4\pi^2 r^3}{g \times M}\). Again, that big \( M \) is my target variable here, so I know what the orbital period is. But again, there's that \( r \) that I don't have, that orbital distance. So in both of these approaches here, both of these equations that I've seen, I ended up with the same unknown variable. I have too many unknowns. So there's got to be something else I can do to solve for this little \( r \) distance. Let's see. What's the one equation I haven't used yet? So, I'm going to go over here and solve for a little \( r \). The one I haven't used yet is this one, the \( v_{\text{sat}} = \frac{2 \pi r}{t} \). So I've got \( v_{\text{sat}} = \frac{2 \pi r}{t} \). Now in this situation for this equation, I have two knowns and only one unknown. So now I can use that to solve for my little \( r \). So let's see. I've got, moving everything over to the left side. I'm going to get \( v_{\text{sat}} \times t \div 2 \pi \) is going to give me little \( r \). So that means that that little \( r \) here, if you just go ahead and plug everything in, is going to be, let's see, 75,100 times the period. Now, that \( t \) is expressed in hours. So first I need to multiply by 3,600 to get into seconds. If you do that, you should get \( 1.008 \times 10^5 \), that's seconds. So now I'm just going to go ahead and plug that in. \( 1.008 \times 10^5 \). Now, I'm just going to divide by \( 2 \pi \). So, if I divide by \( 2 \pi \), I'm going to get the orbital distance, that little \( r \). That's \( 1.2 \times 10^8 \). So now what I can do is now, I can take this little \( r \) that I found and I can plug it back into either this \( t^2 \) equation because now I only have this unknown, or I can plug it back into the \( v_{\text{sat}} \) equation so that I still only have one known. So, really the choice is up to you. I'm going to just keep going with the \( v_{\text{sat}} \) approach. So I've got now my little \( r \) distance. So I've got \( v_{\text{sat}} = \sqrt{\frac{g \times M}{r}} \). So now I'm just going to go ahead and solve for that big \( M \) because I already have everything else. Right? So I've got \( v_{\text{sat}} \) and I'm going to square both sides because I want to get rid of the square root. So I've got \( v_{\text{sat}}^2 = \frac{g \times M}{r} \), and then I just move the \( r \) over everything over to the left side. So I get \( v_{\text{sat}}^2 \times r \div g = M \). And if you plug all that stuff in, what you're going to get is 75,100 squared times the radius, 1.2 times \( 10^8 \), and then we've got to do divided by the gravitational constant, so times \( 10^{11} \). And if you do that, you should even get the mass of the planet which is \( 1.01 \times 10^{26} \) and that's in kilograms. That's it for this one. Let me know if you guys have any questions with that.

A distant planet orbits a star 3 times the mass of our Sun. This planet of mass 8 × 10²⁶ kg feels a gravitational force of 2 × 10²⁶ N. What is this planet's orbital speed and how long does it take to orbit once?

• A. 1 × 10⁵m/s; 2.5 × 10⁶s
• B. 1.29 × 10⁵m/s; 2.5 × 10⁶s
• C. 1 × 10⁵m/s; 1.94 × 10⁶s
• D. 1.29 × 10⁵m/s; 1.94 × 10⁶s

Here’s what students ask on this topic:

What is the formula for the orbital speed of a satellite?

The formula for the orbital speed of a satellite in a circular orbit is given by:

$\sqrt{\frac{GM}{r}}$

where $G$ is the gravitational constant, $M$ is the mass of the planet (e.g., Earth), and $r$ is the orbital radius (distance from the center of the planet to the satellite). This formula ensures that the satellite maintains a stable circular orbit.

How do you calculate the orbital period of a satellite?

The orbital period $T$ of a satellite is the time it takes to complete one full orbit around the Earth. It can be calculated using Kepler's third law:

$T^2=\frac{4{\pi }^2{r}^3}{GM}$

where $G$ is the gravitational constant, $M$ is the mass of the planet, and $r$ is the orbital radius. This equation shows that the period increases with the cube of the orbital radius.

Why does the orbital speed of a satellite decrease as the distance from Earth increases?

The orbital speed of a satellite decreases as the distance from Earth increases because the gravitational force exerted by the Earth weakens with distance. The formula for orbital speed is:

$\sqrt{\frac{GM}{r}}$

As the orbital radius $r$ increases, the value of the denominator increases, leading to a decrease in the orbital speed. This is because a satellite farther from Earth requires less speed to counteract the weaker gravitational pull and maintain a stable orbit.

How do you find the height of a satellite above Earth's surface?

To find the height $h$ of a satellite above Earth's surface, you first need to determine the orbital radius $r$ using the orbital speed formula:

$r=\frac{GM}{v^2}$

Once you have $r$, you can find the height $h$ by subtracting the Earth's radius $R$ from the orbital radius:

$h=r-R$

This gives you the height of the satellite above the Earth's surface.

What is Kepler's third law and how is it used in satellite motion?

Kepler's third law states that the square of the orbital period $T^2$ of a planet is directly proportional to the cube of the semi-major axis $r^3$ of its orbit:

$T^2=\frac{4{\pi }^2{r}^3}{GM}$

In satellite motion, this law helps relate the orbital period to the distance from the central body (e.g., Earth). It is used to calculate the time a satellite takes to complete one orbit, given its distance from the planet, and vice versa.