Elliptical orbit changing as a star's mass increases

Question

I'm studying Kepler's Laws, specifically the orbit of the Earth around the Sun. I know that if the Earth was more massive, the orbit would not be significantly affected. If the Sun was more massive, I know the velocity of Earth's orbit around the Sun would increase, but how would the shape of the orbit change? This is a theoretical case, neglecting conservation of mass and assuming the Sun's radius and volume don't change.

Answer 1

I think the answer is rather complicated and depends on (i) the timescales of the mass gain; (ii) the initial eccentricity; (iii) the position of the object in its orbit when the mass gain occurs.

The case of "adiabatic" mass changes is dealt with in Veras et al. (2011). In this instance, the mass loss timescale is much longer than the orbital period. If this is the case then they provide an analytic solution of the form $$ e(t) = e_0 - \frac{P \dot{M}}{2\pi \mu} \frac{\left(1-e_0^2\right)^{3/2}\sin f}{1 - e_0 \cos f},$$ where $f$ is the true anomaly, $e_0$ is the initial eccentricity, $\dot{M}$ is the time derivative of the stellar mass and $\mu = (M+m)$ (the total mass of star and planet). For a positive $\dot{M}$ the eccentricity decreases with time as the orbit shrinks.

If however you mean, assume that some parameters of the Sun-Earth system are kept the same whilst others are instantaneously changed, then that would be different. You need to specify what is being kept constant -- energy, angular momentum, both?

I'll assume you want to keep angular momentum constant. Thus if $M_1 \gg M_2$, we can say the angular momentum before the mass gain $$ J_0 = M_2 (a_0[M_{1,0} + M_2](1 -e_0^2))^{1/2} \simeq M_2 (a_0 M_{1,0}(1 -e_0^2))^{1/2}$$ and this equals the angular momentum after the mass gain $$ J = M_2 (a[M_1 + M_2](1 - e^2))^{1/2} \simeq M_2 (a M_1(1 - e^2))^{1/2} $$ Here, $M_{1,0}$ is the original stellar mass and $M_1$ is the new, larger stellar mass; $e_0$ is the original eccentricity and $e$ is the new eccentricity; $a_0$ is the original mean separation and $a$ the new mean separation. Equating these, we have $$1 - e^2 = (1-e_0^2) \left[\frac{a_0 M_{1,0}}{a M_1}\right] \tag{1}$$

To work out what happens to the eccentricity we need to work out what happens to the trailing factor.

The energy equation for an elliptical orbit tells us that $$ -\frac{GM_{1,0}M_2}{2a_0} = -\frac{GM_{0,1}M_2}{r} + \frac{1}{2}M_2 V^2 \tag{2}$$ where $V$ is the instantaneous orbital speed at instantaneous separation $r$ and again I assume $M_2 \ll M_{1,0}$.

A similar equation applies after the mass gain: $$ -\frac{GM_{1}M_2}{2a} = -\frac{GM_{1}M_2}{r} + \frac{1}{2}M_2 V^2 \tag{3}$$ where we assume the same instantaneous speed and position.

Using equation (2) to get an expression for $r^{-1}$ and substituting this into equation (3), I get $$ \frac{1}{a} = \frac{1}{a_0} +\frac{V^2}{GM_{1,0}} - \frac{V^2}{GM_1} \tag{4}$$

This tells us that as the mass is increased then $a<a_0$ and the orbit shrinks. Substituting equation (4) into equation (1) $$ 1- e^2 = (1-e_0^2)\left[ \frac{a_0 V^2}{GM_1}\left(1 - \frac{M_{1,0}}{M_1}\right) + \frac{M_{1,0}}{M_1}\right] \tag{5}$$

For a circular orbit, $e_0=0$, $a_0 V^2/GM_{1,0}=1$ and $$ e^2 = 1 - \left[ \frac{M_{1,0}}{M_1}\left(2 - \frac{M_{1,0}}{M_1}\right) \right]$$ e.g. If $M_{1,0}/M_1= 0.5$ (the mass doubles), then $e=0.5$.

For other initial eccentricities, the result depends on the instantaneous value of $a_0 V^2/GM_{1,0}$, which can be bigger or smaller than 1. So if the mass change takes place with the Earth as perihelion, than $a_0 V^2 = GM_{1,0} (1+e_0)/(1-e_0)$ and $$ 1- e^2 = (1-e_0^2)\left[ \frac{M_{1,0}}{M_{1}}\left(\frac{1+e_0}{1-e_0}\right)\left(1 - \frac{M_{1,0}}{M_1}\right) + \frac{M_{1,0}}{M_1}\right] \tag{6}$$

If the mass loss occurred at aphelion $$ 1- e^2 = (1-e_0^2)\left[ \frac{M_{1,0}}{M_{1}}\left(\frac{1-e_0}{1+e_0}\right)\left(1 - \frac{M_{1,0}}{M_1}\right) + \frac{M_{1,0}}{M_1}\right] \tag{7}$$

So for concrete examples - let's increase the the mass of the Sun by 10%, for the Earth with $e_0=0.017$. Using equation (6), if this occurs at perihelion then $e = 0.084$, at aphelion $e=0.100$.

Answer 2

If the existing orbit were already circular then any changes like you describe would immediately result in an elliptical orbit. After that the orbit would tend toward a circular orbit. How long that takes depends on the total tidal forces. A rigid body would take longer than a pliable one as tidal forces are exchanged.

Answer 3

Kepler's laws are not correct:

No area law, no elliptical orbits, nor period. Starting with Newton's mechanical laws, Kepler's area law does not exist. Newton's universal attraction force is radial. $F=F_r$. A side force component, a perpendicular force does not exist $F_p=\frac {m dV_p} {dt}=0$. Then when integrated , we get $V_p=Ct$ . Kepler says $r V_p=area=Ct$. Newtons equation is a prove. Kepler's area law is an estimation.

When using $V_p=Ct$ in the differential form of the energy conservation equation,we get

$r=-4*{t^2}+4 t T-4 \frac {T^2} 6 $ as the equation of celestial motion. This is a spiraled orbit and not an ellipse.

When to the validity of the period law: Newton says period law is valid for circular motion with non accelerated velocity. Kepler says period law is valid even for elliptical orbits with accelerated velocity. High velocity at perihelion low , velocity at aphelion. And no body feel this acceleration?