# Calculation of the speed of the ISS from the circulation time

This is my first question, and as a chemist, my physics vocabulary is a bit limited. A younger friend of mine asked me about a physics question from school, but I am not able to solve it, and I think the question is not providing enough information. The topic is "Kepler's Law."

The question says:

The circulation time of the ISS is 91 minutes. Which height above ground does the ISS have? What is its speed?

First, I looked up Kepler's law, but to use the third law I would need something else circulating around the earth like the moon and information on its semi-major axis.

Then, I found Newton's gravitation law which needs the masses of the two objects.

So maybe you have some other ideas that I should look up to find an answer how to calculate the semi-major axis of the ISS just from the circulation time without using google for the circulation time of the moon or the mass of the earth.

• "...like the moon" Could you look up its distance and period? Aug 24 at 19:21
• well the moon takes 1 month by definition to go around. Minus 1/12 for the sidereal correction. Don't need to look that up.
– JEB
Aug 25 at 0:28
• I don't understand the aversion to googling the information you need (mass/radius of Earth). The purpose of the question is to test your understanding of physics, not to inform the examiner of the height of the ISS. Aug 25 at 11:57
• @OscarBravo I think the assumption is that a school problem usually has all the information needed in it. If you're meant to use outside sources, you could just google the the altitude and velocity of the ISS right away.
– noah
Aug 25 at 12:00
• Thank you noah, that is exactly what it is about. You are supposed to learn the formula and use it accordingly not looking up information on the internet. It might be different but in germany you are almost never supposed to search for given numbers outside a standard book and then it is explictly mentioned or discussed in class before. Aug 25 at 12:11

Which height above ground does the ISS have?

The wording of this question makes me think that the question assumes the orbit to be circular, since an ellipse will not have a constant height above ground.
Also, since this is at the level of school teaching, I think they are just assuming a circular earth and a circular orbit around the earth.

If that is the case, then you can just use the equation

$$\frac{mv^2}{h+r} = \frac{GmM}{(h+r)^2}$$

where $$h$$ is the height above ground,
$$r$$ is the radius of Earth,
and $$M$$ is the mass of Earth.
For $$v$$, you can substitute the perimeter of the circular orbit / circulation time i.e. $$v=\frac{s}{t}=\frac{2\pi\,(r+h)}{91\text{ minutes}}$$ Take care to convert to same units.

This will give you an equation with only one unknown, $$h$$.
You can solve it to get $$h$$.

Of course, in the real world a lot of these assumptions will not hold. But, I think the school question is going for a more simplistic model with those simplified assumptions.

• This still means, I need to look up the radius and the mass of the earth. Then I could google the height of the ISS as well. Aug 25 at 11:42
• @Inselino Radius and mass of the earth are values that are generally considered as known/given values when studying mechanics etc. I do not think it is the same thing as just looking up the answer. You have mentioned elsewhere about how it is in Germany, but where i come from, those were known values that could be used in calculations as and when required
– user312165
Aug 25 at 12:55
• But this is 9th grade in school not studying mechanics. Aug 25 at 13:52

Kepler's third law relates the period of an orbit to its radius, using only constants like $$G$$ and the mass of the body about which the object is rotating (here: the Earth). This will directly give you the result with information that is readily available. You are correct that, if you wanted to solve that without the mass of the Earth, you could use another set of equations with the same proportionality constant, such as those from the moon.

Then, once you have radius and orbital period, speed should be straightforward to calculate by using the relationship between the radius and circumference of an orbit.

• I wanted to know if I can solve it without using google. Mass of the earth is not provided in the question. Aug 24 at 19:47
• It's probably something that the text the question comes from assumes can be looked up. Often that's in a table at the front or back of an intro physics book. Aug 24 at 20:02
• There's always a missing variable that is easy to look up: radius of the earth, mass of the earth, etc Aug 24 at 22:32
• Yeah but in this case, I could look up the height of the ISS as well. But I found the answer to the question online, the teacher missed out on some information. Aug 25 at 11:42

You know the acceleration g at the radius of the earth. you know the centipededal acceleration of ISS, you can calculate the velocity at the radius of the earth (in vacuum) then you can use Kepplers law to calculate the radius of ISS and therefor the high as diference

thanks for all the answers. As I could notice from the answers, there is NO way without getting more information like the mass of the earth or the distance earth-moon. But if you use google anyway, you could just look up the distance above ground of the ISS. So this is not the solution.

But we looked up the exact question (originaly in german) in the internet and we found out, that there is this exact question. But it originates from a sheet with a lot of questions and a lot of additional variables like the circulation time and the distance of the moon.

In this case, the teacher just copy pasted the question without the additional information provided in the original source.

• Please move this additional insight as a new section to the question. Use formatting options to structure it in a clear way as addition. Answers should only be used for actual answers. Aug 25 at 12:19
• But this is the answer. There is no way without additional information. Aug 25 at 12:31