# How fast do you need to go to orbit the earth?

Charity coin collectors at the shopping centre remind me that the orbital speed depends on the altitude: the coin orbits slowly at high altitude, then faster as it goes down the drain.

So how fast would you need to go at the surface of the earth to be orbital? For the question, ignore air and buildings…

17,672 mph

~Mach 23
Faster than a speeding bullet? Much faster than a speeding bullet.

If Jeff Bezos and Richard Branson had gone into orbit, that’s how fast they would have been going on landing (… for the sake of the question, ignoring air and buildings…)

Speed of sound keeps going down as you go higher in altitude. So when you say “Mach 23”, do you mean the speed of the rocket / speed of sound at Sea Level ?

Mach numbers are conventionally quoted assuming dry air, mean sea level, and an atmospheric temperature of 15ºC.

But don’t stress about it. The key figure here is not Mach 23, but 28,440 km/h, the orbital speed for a theoretical orbit at the level of the earth’s surface at the equator.

At that speed, our theoretical orbiting object (if not vapourised by friction) would complete an orbit in just under 1 hour 25 minutes. Be sure not to stand it its way.

May i have a cite for that ? I am used to fluid dynamics / flow calculations, where the Mach number is always for the prevalent conditions.

This NASA website has an online calculator, that gives altitude accounted MACH numbers for the same speed of a rocket. https://www.grc.nasa.gov/www/k-12/rocket/mach.html

This is fun stuff - no stress here.

As it happens, the atmosphere is so thin that you can pretty much do that by going just a couple hundred miles up. The ISS orbits at nearly the same speed as the hypothetical surface orbit.

Yeah, low-earth orbit (where the ISS and most satellites orbit) is only very slightly different from being exactly on the surface.

Fun fact: For a surface orbit around a uniform, spherical body (such as any planet), the time it takes for one orbit depends only on the density of the body you’re orbiting. Since many planets and planet-like objects have similar density, they’ll also have a similar orbital period.

I just wondered how fast it was. So … faster than a very fast bullet. Faster than a very fast airplane.

And also,
. slow enough that it would go away, and wouldn’t be back again until you’d had a chance to sit down and think about it,
. slow enough to warn people,
. slow enough that you could watch the track on live video,
. slow enough that a phone call would be quicker than the trip to London

an inordinately pleasing hypothetical.

Relevant short story:

https://www.gutenberg.org/files/32360/32360-h/32360-h.htm

The story @DemonTree linked is “The Holes Around Mars”, for anyone interested in such things. Though it’s not at all clear that the objects in that story behave according to the rules of orbital mechanics as we know them, and in fact there are some strong indications that they don’t.

17,400 MPH is the figure I’ve heard quoted for LEO - which, yes, is pretty close to Telemark’s cite of 17,672 at the earth’s surface.

For a geosynchronous orbit - that is, one for which the orbital period is 24 hours - the altitude is about 22,200 miles, and the orbital velocity is more like 7,000 MPH.

we can calculate such “how fast do you need to go” figures from Kepler’s laws. In this case, look up the gravitational parameter of the Earth to be \mu= 3.986 × 1014 m3/s2, and, for a circular orbit,

v^2 = \frac{\mu}{r}

Therefore the required speed is inversely proportional to the square root of the distance.

Aw Man, beat me to it!

I should have gotten up earlier.

And in case anyone’s wondering, the “gravitational parameter of the Earth” just means the Earth’s mass times Newton’s constant of gravitation. It’s used here because, first, it’s what’s directly useful in the formulas, and second, it can be measured to much greater precision than either m or G in isolation.

Another fun fact. The time for a complete orbit at ground level or low earth orbit is about 90 minutes. This is the same time it would take for a frictionless stone to fall through a hole going through the center of the earth to come back again. This is not a coincidence. By decomposing an orbiter’s motion into tangential and normal (perpendicular to the surface) components you see it is really the same question.