# Orbital height without the gravitational constant and the mass of either object [closed]

Is it possible to calculate the orbital height of an object using only the orbital period T = 93min, radius of the Earth R = 6378km and the gravitational acceleration at the surface of the Earth g = 9,8m/s^2? Until now I've used the equation:

Fc = Fg

mv^2/r = GmM/r^2

v^2 = GM/r

4pi^2r^2/T^2 = GM/r

r^3 = GMT^2/4pi^2

r - distance between the centres of mass of the two objects, G - gravitational constant, M - mass of the Earth, T - orbital period.

But without knowing the gravitational constant or the mass of the Earth I can't seem to calculate this. I believe I have to somehow use the gravitational acceleration, but it only applies to the surface and is very different at the altitude of the satelite. How would I be able to calculate this?

• with the answer of Neil I got the ISS hight 419 km , which is correct, I wonder how you solved this problem with the answer of mike , because g(r) is unknown?
– Eli
Nov 19 at 17:22
• @Eli Could you please tell me how you solved it, I tried asking Neil, but he wasn't of much help Nov 20 at 7:41
• @Eli I can't use M or G. Nov 20 at 8:32
• $R+h=\left( \dfrac{GMT^{2}}{4\, \pi^2}\right) ^{\dfrac{1}{3}}.=\left( \dfrac{gT^2R^{2}}{4\,\pi^2 }\right) ^{\dfrac{1}{3}}$
– Eli
Nov 20 at 9:19

First get $$\omega$$ from, $$\omega=\frac{2\pi}{T}$$, then equate it with,

$$\omega^2(h+r)=\frac{GM}{(h+r)^2}\Rightarrow (h+r)^3=\frac{GM}{\omega^2}$$

This is kepler's third law, where $$h$$ is height of ISS from earth's surface, $$r$$ is radius of earth, $$M$$ is mass of earth.

• Could you please elaborate a little, I am still a student and don't have a very deep understanding of physics. What would be the final equation for the orbital height? Until now I was using that the centrifugal and gravitational force are equal but act in opposite directions, so I can set them equal to one another and solve for r, which gave me: r^3 = GMT^2/4*pi^2. Nov 19 at 22:10
• @JanHrubec Use $r+h$ instead of $r$, then minus $r$ from it to get $h$. . Nov 19 at 23:23
• Ok, but I still don't know the value of G or M so I can't calculate it Nov 20 at 7:38
• @JanHrubec Okay, $GM=gr^2$. Now either put it wherever it needs or reformulate above equation by taking first order expansion as $r>>h$. Therefore, $1+\frac{3h}{r}=\frac{1}{\omega^2r}$ or $h=\frac{1}{3\omega^2}-\frac{r}{3}$. Nov 20 at 7:53
• Amazing, could I just ask, is GM=gr^2 based in any equation or is there any way I could prove it? Nov 20 at 8:19

You do not need $$G$$. Just use proportionality: The gravitational acceleration at height $$r$$ is related to the $$g$$ at the surface $$R_{\rm earth}$$ by $$g(r) = g_{\rm surface} \frac{R_{\rm earth}^2}{r^2}$$

• $r\mapsto R_e+h$
– Eli
Nov 19 at 16:46
• @mike stone g(r) is unknown, so how you solve your equation for h ?
– Eli
Nov 19 at 16:52
• @Eli Fg = g(r)*m, and you know everything except for m and r, but m cancels out and r is the only unknown, so the final equation is r = cube root(gR^2T^2/4pi^2) where g is the gravitational acceleration on the surface (9,8m/s^2), R is the radius of Earth (6378000m) and T is the orbital period (5880s) Nov 20 at 9:44
• @mike stone Thank you very much, this has worked, but could you explain why the relationship actually works? Nov 20 at 10:48
• It is just propertionality from Newton's inverse square law: $g_{\rm gravity}=GM/r^2$. Take ratios and the $GM$ factor cancels. Nov 20 at 13:45