Gravity effects when deleting stars or planets

Question

In one of the latest SCIFI movies, they collect a star absorving it entirely in a kind of planet size gun/machine. As mass cannot disappear, I assume it was just concentrated inside their machine (until fired).

1. As the mass is displaced, what would be the effects on the orbit of other bodies in the system without that star?

2. Would it change its center to the Death Star?

3. The same when an entire planet is destroyed, can we expect huge changes in other planets orbits?

4. What would happen in our solar system, if the Sun got displaced or simply disappear?

5. Would all planet orbits collapse?

6. Or would planets continue moving until captured by the gravity of other bodies?

In films this is shown just like a gun shot, a single event that ends with destruction of a star or planet. I'm wondering if the quick change in mass and orbits would not produce a huge destabilizing effect in the entire system.

Answer 1

The scenario you suggest is of course hypothetical, but in all cases you must conserve angular momentum and mass/energy.

So for example: If you have a way of removing mass from a star in such a way that the mass disappears outside the orbit of the planets (in astrophysics this is accomplished simply by mass loss - either the star has a wind that expels mass outwards, or something more catastrophic like a supernova) then the mass of the central object in that system has obviously decreased, but the angular momentum of the planets will not have changed.

Kepler's third law tells us that $$ \frac{a^3}{P^2} \simeq \frac{GM}{4\pi},$$ where $a$ is the semi-major axis, $P$ is the orbital period and $M$ is the mass of the star we are assuming that the star is much more massive than the planets.

Assuming circular orbits for a moment, the angular momentum is $L = 2\pi m a^2/P \propto m a^{1/2} M^{1/2}$, where $m$ is the (constant) planet mass. Thus, if $M$ suddenly decreases, then $a$ must increase to compensate. So the planets move into wider orbits.

However, if the mass is lost on a timescale shorter than the orbital period of the planet, it's orbit would likely become eccentric. The details are given in Elliptical orbit changing as a star's mass increases

In the limit that the mass was completely removed from the system, then $a$ tends towards infinity and the planets would start travelling on roughly straight trajectories (though still influenced by the gravitational pull of the other planets) with the initial velocities that they had when the star was removed. The finite speed of light means that the removal of the star would not be "felt" by the planet until the light travel time between the original position of the star and the planet had elapsed.

Taking the solar system as an example. If the Sun magically disappeared, it would be around 8 minutes before the Earth started to move in a straight line trajectory at about 30 km/s with respect to where the Sun was; a tangent to its original orbit. The velocity of the Earth with respect to the Sun (and with respect to the other planets) is large enough that it exceeds any escape velocity with respect to the other planets, so could not be captured (for instance) by Jupiter.

Answer 2

If at any time the speed of the planet in the reference frame of the star exceeds the escape velocity $\sqrt{2GM_\star/r}$, where $M_\star$ is the mass of the star and $r$ is the distance from the star to the planet, it will escape in a hyperbolic trajectory (or straight line if $M_\star\rightarrow0$).

As noted in the other answers, the result of the scenario will depend on the exact way that your central star is (re-)moved. I wrote a small Python code that calculates the orbit of a planet orbiting a $1\,M_\odot$ star (assuming $M_\mathrm{planet} \ll M_\star$). The simulation uses a Leapfrog integration scheme for advancing the position of the planet in time. After a time $t_\mathrm{go}$, you can move the star by a distance $dx$ per time step, and/or you can shrink the mass of the star by mass $dM$ per time step. You can play around with the exact values to match your favorite scenario.

The figure shows a simulation where after 10 years the star starts to move $dx = 0.1\,\mathrm{AU}$ and shrink by $dM = 0.005\,M_\odot$ per time step $dt = 0.25\,\mathrm{yr}$. With these parameters, after 51 years the planet escapes its star.

^{Red and yellow lines show the trajectories of the planet and the star, respectively.}

import numpy as np
from matplotlib import pyplot as plt

G  = 6.67384e-8                                 #Gravitational constant
Ms = 1.989e33                                   #Solar mass
M  = 1 * Ms                                     #Stellar mass
AU = 1.496e13                                   #Astronomical unit
yr = 365.25 * 86400                             #Year in seconds

xo = np.array([0,0])    * AU                    #Star initial position
x  = np.array([5.2,0])  * AU                    #Planet initial position
v  = np.array([0,13.1]) * 1e5                   #Planet initial velocity

moveStar = True
dx = np.array([.1,.1])  * AU                    #Star displacement per time step
shrinkStar = True
dM = .01 * Ms

dt   = .25 * yr                                 #Time step
tgo  = 10 * yr                                  #Time that star starts moving
tfin = 25 * yr                                  #Time that simulation stops

r = np.sqrt((x[0]-xo[0])**2 + (x[1]-xo[1])**2)      #Planet-star distance
a = -G*M / r**2 * (x-xo)/np.linalg.norm((x-xo))     #Acceleration
t = 0.                                              #Time start

plt.clf()
plt.ylabel('x / AU')
plt.xlim([-6,6])
plt.ylim([-6,6])
plt.scatter(xo[0],xo[1],c='y',edgecolor='y',s=250)  #Star symbol

while t <= tfin:
  plt.xlabel('x / AU;   t = {:.1f} yr'.format(t/yr))
  plt.xlim([xo[0]/AU-6,xo[0]/AU+6])
  plt.ylim([xo[1]/AU-6,xo[1]/AU+6])
  plt.scatter([x[0]/AU],[x[1]/AU],c='r',edgecolor='r',s=10,alpha=0.25)

  #Leapfrog:
  xold = x                                          #Old position
  t    = t + dt                                     #Update time
  x    = x + v*dt + .5*a*dt**2                      #Update position
  aold = a                                          #Old acceleration
  a    = -G*M / r**2 * (x-xo)/np.linalg.norm((x-xo))#Update acceleration
  v    = v + .5*(a+aold)*dt                         #Update velocity
  r    = np.sqrt((x[0]-xo[0])**2 + (x[1]-xo[1])**2) #Update position

  plt.plot([x[0]/AU,xold[0]/AU], [x[1]/AU,xold[1]/AU],'r',alpha=.25)
  plt.draw()
  plt.pause(0.05)

  if t/yr > tgo/yr:
    if moveStar:
      plt.scatter(xo[0]/AU,xo[1]/AU,c='white',edgecolor='white',s=300)
      xo = xo + dx                              #Move star
      plt.scatter(xo[0]/AU,xo[1]/AU,c='y',edgecolor='y',s=250)
    if shrinkStar:
      plt.scatter(xo[0]/AU,xo[1]/AU,c='white',edgecolor='white',s=300)
      M = max(M-dM,0.)                          #Shrink star
      plt.scatter(xo[0]/AU,xo[1]/AU,c='y',edgecolor='y',s=250*(M/Ms))

Answer 3

Your final question very much correlates with a famous thought experiment.If the Sun was suddenly removed the planet s will still continue to stay in orbit. For 8 minutes and 20 seconds. This is because the speed of the space time fabric or simply putting gravity travels at the speed of light. That is, the earth will be devoid of sunlight and will move tangentially to the orbit in the same time,ie an avg of 8mins 20seconds after the Sun disappeared.