# Can the mass of an orbiting object and the object being orbited be determined by the distance and orbit velocity alone? [duplicate]

This question is the inverse of: "Could an object orbit while moving at twice the speed, but at the same distance, if it had half the mass?"

I'm curious about the nature of orbits, but am not well enough versed in mathematics to understand Kepler's laws well. I have been wondering if the mass of a planet and a star it orbits could be determined based solely on the distance and speed of the orbit, or if the ability to orbit at a given speed/distance was based relatively on the mass of both objects (i.e., we could determine the ratio of the mass of the two objects, but not the actual mass).

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## marked as duplicate by Waffle's Crazy Peanut, Chris White, Qmechanic♦Jul 1 '13 at 19:03

No, if you only know the distance between the objects and the relative orbital velocity of the planet, you cannot determine its mass. In fact, if you only know the distance and velocity at one particular moment, you don't have enough information to determine the orbit.

Suppose we know the distance $r$ and the relative orbital velocity $\vec{v}=(v_r,v_T)$ of a planet at a given moment. Here, $v_r=\dot{r}$ is the radial velocity component, and $v_T$ the tangential component. The orbit of the planet has two constants of motion: the specific orbital energy $E$ and the specific relative angular momentum $h$: \begin{align} E &= \frac{1}{2}v_{r}^2 + \frac{1}{2}v_{T}^2 - \frac{\mu}{r}= -\frac{\mu}{2a},\\ h^2 &= r^2\,v^2_{T} = \mu a(1-e^2), \end{align} where $a$ is the semi-major axis of the orbit, $e$ is the orbital eccentricity, and $\mu = G(M_\text{p} + M_\text{s})$, with $M_\text{p}$ the mass of the planet and $M_\text{s}$ the mass of the star. So we have two equations and three unknowns $(\mu,a,e)$, in other words we need additional information to solve them.

For instance, if we assume that the orbit is circular, then $e=0$, and we can solve for $\mu$ and $a$. Another possibility is that we know the distance and velocity at two instances $t_1$ and $t_2$, then $$\frac{1}{2}v_{r}^2(t_1) + \frac{1}{2}v_{T}^2(t_1) - \frac{\mu}{r(t_1)}=\frac{1}{2}v_{r}^2(t_2) + \frac{1}{2}v_{T}^2(t_2) - \frac{\mu}{r(t_2)},$$ and we can solve for $\mu$, and, from the first two equations, we know $a$ and $e$. A third possibility is that we know the orbital period $T$ of the planet. In that case, we can use Kepler's Third Law: $$T^2 = (2\pi)^2\frac{a^3}{\mu},$$ which, in combination with the first two equations, yields $(\mu,a,e)$. In any case though, we can only derive $\mu = G(M_\text{p} + M_\text{s})$, i.e. we only know the sum of the masses (which will be dominated by the mass of the star).

If you want to derive the mass of the planet, you need to know the motion of the planet and the star with respect to their common centre of mass. If $(r_\text{p},v_{r,\text{p}},v_{T,\text{p}})$ and $(r_\text{s},v_{r,\text{s}},v_{T,\text{s}})$ are the position and velocity of the planet and the star with respect to their common centre of mass, then \begin{align} E_\text{p} &= \frac{1}{2}v_{r,\text{p}}^2 + \frac{1}{2}v_{T,\text{p}}^2 - \frac{\mu_\text{p}}{r_\text{p}}= -\frac{\mu_\text{p}}{2a_\text{p}},\\ h^2_\text{p} &= r^2_\text{p}\,v^2_{T,\text{p}} = \mu_\text{p} a_\text{p}(1-e^2),\\ E_\text{s} &= \frac{1}{2}v_{r,\text{s}}^2 + \frac{1}{2}v_{T,\text{s}}^2 - \frac{\mu_\text{s}}{r_\text{s}}= -\frac{\mu_\text{s}}{2a_\text{s}},\\ h^2_\text{s} &= r^2_\text{s}\,v^2_{T,\text{s}} = \mu_\text{s} a_\text{s}(1-e^2), \end{align} and in addition, if we know $T$, $$T^2 = (2\pi)^2\frac{a^3}{\mu} = (2\pi)^2\frac{a^3_\text{p}}{\mu_\text{p}} = (2\pi)^2\frac{a^3_\text{s}}{\mu_\text{s}}.$$ In principle, these are five equations with five unknowns $(\mu_\text{p},\mu_\text{s},a_\text{p},a_\text{s},e)$ (actually, these are six equations, but they're not independent; also note that the eccentricities of the orbits are the same). Also \begin{align} r &= r_\text{p} + r_\text{s},\\ a &= a_\text{p} + a_\text{s},\\ M_\text{p}r_\text{p} &= M_\text{s}r_\text{s},\\ M_\text{p}a_\text{p} &= M_\text{s}a_\text{s},\\ \mu_\text{p} &= \frac{GM_\text{s}^3}{(M_\text{p} + M_\text{s})^2},\\ \mu_\text{s} &= \frac{GM_\text{p}^3}{(M_\text{p} + M_\text{s})^2}. \end{align} Once the equations are solved, you can derive $a$ and $\mu$ (using Kepler's Third Law again). So you have $M_\text{p}/M_\text{s}$ and $M_\text{p} + M_\text{s}$, so that you can derive $M_\text{p}$ and $M_\text{s}$ separately.

In practice, it is usually too difficult to measure $r_\text{p}$ and $r_\text{s}$, because the centre of mass will be very close to the centre of the star. But by measuring the velocities at two instances $t_1$ and $t_2$, we can treat $r_\text{p}$ and $r_\text{s}$ as additional unknowns and calculate them as well.

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If you know the distance between the two objects and the velocity of the orbiting object, you can only determine the mass of the object being orbited. So, in the case of the Earth and the Moon, if you know the distance between the two and the velocity of the Moon, you can determine the distance the Moon travels in one orbit and how long it takes - that is, the period. From the period of the orbit you can determine the mass of the Earth, but not the mass of the Moon.

So the answer to the title question is you can only determine the mass of the object being orbited. Regarding the question posed in the first paragraph of the text, the mass of the object that is orbiting does not affect how 'fast' it orbits the other object.

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"you can only determine the mass of the object being orbited". That is strictly speaking not correct. One can only determine the reduced mass of the two objects. – Johannes Jul 1 '13 at 15:24
What @Johannes said. Whether object 1 orbits object 2 or vice versa is just a matter of what reference frame you pick. Considering the barycentric frame leads to the reduced mass. – Kyle Oman Jul 1 '13 at 15:27
Sorry, mistake: only the sum of the two masses can be determined. In any case, the key point to note is that if only one mass parameter can be determined, it has to be a mass parameter that is symmetric in the two masses. – Johannes Jul 1 '13 at 15:31