Calculating the Speed to Get to Low Earth Orbit (and Other Calculations)

Image: Rhett Allain. Calculating orbital velocity.

I was rather disappointed when I did an internet search for “speed to low earth orbit” and didn’t get an answer right away. Fine, let’s fix that. I’m going to go through all the calculations for orbits and stuff. The answers will be right here at the top with the calculations (and more) below.

  • What speed do you need to get to low Earth orbit starting from the surface of the Earth (ignoring air resistance)? Answer: 8133 m/s (18,136 mph)

That’s all your answers right there. If answers are all that you love, then that’s what you shall receive (Princess Leia said that). Let’s get to the calculations.

What is Low Earth Orbit?

I’m not going to give a full physics lesson — that would take too long. Instead, I’ll assume you already know some basic stuff. So, I will start with an object moving around a planet (it could be any planet) a height h above the surface. Note: for Low Earth Orbit, this height is about 400 kilometer.

Here is a diagram (not to scale).

Illustration: Rhett Allain. An object orbiting a planet. It could be Earth, who am I to say?

The key here is that in order for the object (the red ball) to move in a circular orbit at THAT altitude, it must have some specific velocity. An object moving in a circle as an acceleration vector pointed towards the center of the circle. This is the centripetal acceleration and it has a magnitude of:

Yes, you could describe this in terms of the velocity (v) or the angular velocity (ω) with r being the radius of the circular orbit (r = R + h). Oh, R is the radius of the planet and h is the orbital altitude.

Since there is an acceleration, you need a force. This force is the gravitational force. It is also pulling towards the center of the planet (that’s convenient) and has a magnitude of:

Hopefully, it’s clear that M_p is the mass of the planet and m is the mass of the object. G is the gravitational constant (6.67 x 10^-11 N*m²/kg²). Using Newton’s 2nd Law I can set the gravitational force equal to mass times acceleration.

Of course, I should probably substitute R and h in for r:

I’m just going to leave that as v² for now — when we talk about energy, we need the square of the velocity. Oh, you want numbers? Here is a link to my python calculations. That way, you can change the numbers for different planets and stuff.

This is what it looks like.

But that’s just the calculations AT low Earth Orbit. Now we need to figure out how to get there.

Energy Needed for Low Earth Orbit

I know we want to find the “launch velocity” for LEO, but I think it’s best to start with energy. I’ll start by using the system consisting of the object (I’m just going to call it a ball from now on — it’s a ball in orbit) and the planet (it could be the Earth). That means there will need to be work done on the system to increase its energy and there will be both kinetic energy and gravitational potential energy. Oh, I will call the starting position with subscript 1 to be on the surface of the Earth and subscript 2 to be in orbit.

Here we go. In this first version, I will assume the ball starts at rest on the surface of the planet.

Quick note: for the change in gravitational potential energy, there was a negative sign. I brought that into the expression to switch the order of the terms. OK, now I can just add in the expression for the orbital velocity (v_2) to get:

There are two terms with 1/(R+h). Combining those, gives the following.

Notice that this still has the mass of the ball (m) since we are looking at the total work required to get it into orbit. Now I can just add this stuff to my python calculation (it’s the same link). I’ll use a ball mass of 1 kg, just for simplicity. I can then just divide by 1 kg to get the energy to orbit per kilogram with a value of 33.1 million Joules (3.31 x 10⁷ J). Just for fun, this is about the energy in 1654 iPhone batteries (assuming a phone battery is 20,000 Joules).

Launch Speed to Orbit Altitude

Now suppose you want to literally throw a ball into space — but not orbit. This would be just a ball going up and then coming back down. Also, there are different definitions of “space”, but I’m going to use an altitude of 400 kilometers (you can change the value in the code).

This is fairly simple to calculate. Again, I will use the system consisting of the Earth and the ball — but with just three differences. First, there is no work done on the system (W = 0 J). Second, the ball will start off with some unknown (to be calculated) velocity — v_1. Third, the final velocity (when it reaches its highest point) will be zero (and not the orbital velocity).

This gives the following work-energy equation that can be solved for v_1 (the initial velocity).

Let’s add this to my python calculation. If you start from the surface of the Earth (at rest) you need to throw a ball with a velocity 2715 m/s (just for fun, that’s 6,054 mph.

Launch Speed Needed for Orbital Motion

Do you see how these calculations build on each other? Isn’t that nice? You don’t have to start all over from scratch but instead can just use the stuff from before.

OK, let’s say I want to throw a ball (again, with no air resistance) into orbit. How fast would I have to throw it? The system will be the same as the previous calculation — the ball plus the Earth and there is no work done on the system. The only difference in this calculation is that the final velocity of the ball is not zero but instead it’s the orbital velocity value. But remember that I have an expression for the final velocity in terms of R and h and stuff.

Plugging in my values (for Earth) means that you would need to throw the ball with a speed of 8133 m/s (18,136 mph) to get into orbit.

Animating Orbits

Let’s take a break — for some fun. I want to make an animation of a ball in LEO and also a ball launched from Earth to get into orbit. I’m not going to go into all the details, but here is one of my older video that shows how to make a numerical orbit model in python.

Now we can see the problem with “throwing something into orbit”. If you start from the ground and throw an object, how can it get into a circular orbit? Hint: it can’t. Here is some code that shows two balls. The first ball is launched from the Earth at a 10 degree angle (with the correct throwing speed) and the other ball is in orbit. (code here). Here’s what it looks like.

You can see how close LEO is to the surface of the Earth (compared to the size of the Earth). Also, notice that the yellow ball (thrown from Earth) does indeed orbit the Earth. It’s just not a circular orbit and it goes BELOW the surface of the Earth.

OK, enough playing around — let’s get back to work.

Escape Velocity

The faster you throw a ball, the higher it will go before coming back down. OK, technically you could throw it into orbit at ground level and it wouldn’t go up OR down. But that wouldn’t work on the Earth — you know, because of all that air.

If you throw something fast enough, it will go so high that it will NEVER come back. This is the escape velocity.

We can calculate escape velocity with the same work-energy equation that we’ve been using. In this case, the ball will start with some velocity (v_1) and end with a velocity of zero m/s (just barely stops). But how high does it go? How about an infinite height — but for right now, I’m just going to call this final distance r_2.

That means we have the following work-energy equation.

Now, what happens as r_2 goes to infinity? That term of 1/r_2 clearly goes to zero. That means that the escape velocity will be:

Just to be clear, the escape velocity depends on BOTH the mass of the planet AND the radius. This is because the radius (R) tells you how far from the center you start the motion. As R increases (with a constant mass), it’s easier to “escape”.

Using values for the mass and radius of the Earth gives an escape velocity of 11,176 m/s or 24,923 mph.

Oh! Notice that it doesn’t matter which direction you throw the ball — if it’s going fast enough it will still escape. If you don’t believe me, go change that orbital motion python code and give the ball an initial escape velocity.

Rotating Earth and Mountains

I’m combining these last two items (in case you can’t tell). So, what happens if you take the rotation of the Earth into account for the launch speed? Well, if you throw the ball in an eastward direction then the ball is already moving before you even throw it. Yes, that’s because the Earth rotates and you are on the surface of the Earth rotating with it (I’m assuming that’s the case).

Oh, if you throw the ball in another direction then the rotation of the Earth doesn’t help (or it makes it harder). You have to throw in the same direction that the surface of the Earth is moving. Also, the closer you are to the equator, the faster you are moving. Just imagine you were at the North pole. The rotation of the Earth would just make you rotate in place and not actually move.

This is why the USA likes to launch rockets from Florida. It’s closer to the equator than something like Maine and if you launch to the East you go over the ocean instead of land (so if there’s a problem you won’t crash onto a city).

So, how fast does the Earth rotate? No, it’s not once every 24 hours. That is the length of the solar day. This is the time it takes for the Sun to appear at it’s highest point until its again at its highest point. But while the Earth is rotating, it’s also orbiting the Sun. It actually has to rotate MORE than one revolution to get a complete day. The rotation time (with respect to the background) stars is just 23.9344696 hours. This is called a sidereal day. Let’s just call this time ΔT (for now).

That puts your surface-based velocity at v = s/ΔT where s is the total distance traveled (and ΔT is the rotation time). The distance s is a giant circle around the Earth — but s is only equal to the circumference of the Earth at the equator. At the poles, s is actually zero. Here’s a diagram that might help.

Illustration: Rhett Allain. The circumference of a circle depending on latitude.

If you are at the equator (latitude = 0 degrees), then your circular motion radius is the same as the radius of the Earth. Otherwise, it’s just some other radius (R_e). That means your surface velocity will be:

I can subtract this velocity from my throwing velocity to get a slightly lower speed. If the latitude is 90 degrees, you get zero boost — that’s what I have already calculated. What about the equator? Using the values for the Earth I get a surface speed of 465.1 m/s (1037 mph) and an adjusted throw speed of 7667.6 m/s (17,099 mph).

Now, what about throwing from a mountain? This isn’t too difficult. The only change to the equation is that the starting position will be R+d instead of R (where d is the height of the mountain).

Let’s say that we ignore the rotation of the Earth and make a comparison between a throw from sea-level (d = 0 meters) and the top of Mt. Everest (d = 8849 meters). This would require a throw speed of 8121.99 m/s (18,112 mph). Note, in the code you can change both the starting height AND the latitude — for fun.

OK, hopefully the next time I search for these answers I will find this page.



Physics faculty, science blogger of all things geek. Technical Consultant for CBS MacGyver and MythBusters. WIRED blogger.

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Rhett Allain

Physics faculty, science blogger of all things geek. Technical Consultant for CBS MacGyver and MythBusters. WIRED blogger.