5
$\begingroup$

When we calculate escape velocity for an object from earth, what we do is, we conserve energy. So if $v$ is the escape velocity, we write $$-\frac{GM_em}{R_e}+\frac{1}{2}mv^2=0$$ since we are assuming that the object escapes earth's gravitational field at infinity and the velocity just becomes $0$ at infinity. But there are other celestial objects in the solar system. So,for example,when the sun is considered,won't the escape velocity of earth change as well? Since then the energy conservation will be as follows: $$-\frac{GM_sm}{r}-\frac{GM_em}{R_e}+\frac{1}{2}mv^2=0,$$ where $r$ is the radius of earth's orbit around sun. Here, the object will again escape the earth only at infinity and the velocity also just becomes $0$. Here we can clearly see that escape velocity of earth is changing as other celestial objects are taken into consideration. So why do we still say escape velocity of earth is $11.2$ km/s whereas this value was derived assuming that no other celestial objects other than earth is being considered. Isn't it wrong? I am really confused.

$\endgroup$
3
  • 2
    $\begingroup$ This may help: xkcd.com/681 or it might add to the confusion. ;) $\endgroup$
    – PM 2Ring
    Aug 30, 2023 at 6:00
  • 1
    $\begingroup$ If you include the Sun's gravitational potential in your energy calculation you also need to (somehow) include the kinetic energy due to the Earth's speed relative to the Sun. $\endgroup$
    – PM 2Ring
    Aug 30, 2023 at 6:03
  • $\begingroup$ See space.stackexchange.com/q/3612 $\endgroup$
    – Jeffrey
    Aug 30, 2023 at 13:10

4 Answers 4

9
$\begingroup$

It depends how you calculate the escape speed. 11 km/s is not the speed required to escape to infinity, for the reasons you describe, but it is (approximately) the speed to escape being gravitationally bound to the Earth. Thus launching something at 12 km/s from the Earth means it won't be coming back, but it will still be orbiting the Sun within the Solar System.

We can make the usual escape speed approximation if the other celestial bodies are far enough away that the additional gravitational potential due to them can be considered unchanging on either side of the Earth. In this case, the gravitational potential of the Sun is just an additive background perturbation that applies equally wherever the body is in the system.

If this approximation applies, then the sum of the kinetic energy and change in gravitational potential energy being zero tells us that the object can escape the gravitational potential of the Earth, but not necessarily from the background potential of the Sun in which the Earth is embedded.

Of course, the gravitational potential of the Sun is not constant either side of the Earth, it varies with distance from the Sun. Leaving aside the complication of angular momentum, if you imagine the potential wells of the Sun and the Earth embedded within that; you can probably see that it would be slightly easier (than the simple calculation of escape speed) to escape from the Earth towards the Sun than away from it.

$\endgroup$
6
  • $\begingroup$ I don't actually understand what's wrong with the last equation I made. I mean I don't see any thing wrong in conserving the energy though I must be wrong. Could you please correct the final equation in that case? $\endgroup$
    – a_i_r
    Aug 30, 2023 at 13:42
  • $\begingroup$ @a_i_r There's nothing wrong with it of you want to work out the escape speed required to get out of the Solar System. $\endgroup$
    – ProfRob
    Aug 30, 2023 at 14:44
  • $\begingroup$ @a_i_r But if you do, don;t forget that the Earth already has a velocity with respect to the Sun, which should be added (as a vector) to your launch velocity, prior to calculating the kinetic energy in that case. $\endgroup$
    – ProfRob
    Aug 30, 2023 at 14:58
  • $\begingroup$ Thank you,so in this case,if the launching speed is $v$ for getting out of the solar system with $v_e$ being earth's velocity wrt sun,then the desired equation is : $-\frac{GM_sm}{r}-\frac{GM_em}{R}+\frac{1}{2}m(v_e+v)^2=0$. Am I right? $\endgroup$
    – a_i_r
    Aug 30, 2023 at 15:06
  • $\begingroup$ I think that, because of the way tidal forces work, while it is very slightly easier to escape from the Earth toward the Sun than away from the Sun, the latter is still slightly easier that it would be to escape from the Earth if there was no Sun; but escaping from the Earth if there was no Sun would be slightly easier that escaping from the Earth in a direction perpendicular to the direction to the Sun. $\endgroup$
    – Litho
    Aug 30, 2023 at 15:30
1
$\begingroup$

The escape velocity of 11.2 km indeed does not include the effect of other bodies, such as the Moon, the Sun, the planets, but also the Milky Way etc. It also does not include the rotation of Earth around its axis nor the Sun, nor the rotation of the solar system around the galaxy. The precise escape, or rather required initial, velocity depends on the latitude, the orbit, say around the moon or a planet, that you are aiming for and how much propulsion you are bringing on your rocket. To save fuel you could decide to launch your rocket to the moon from the equator at new moon.

$\endgroup$
0
$\begingroup$

Escape from the Earth is a specific, solvable problem. The general, multibody, problem is much messier. Gravity assist trajectories allow a spacecraft to, in principle, escape the Solar System, the Galaxy, and even the local supercluster with little change in velocity beyond that needed to reach a Lagrange point. From Earth's surface, that's a bit less than what we call "escape velocity". From there, a spacecraft can escape into solar orbit and subsequently gain velocity from encounters with planets. This can lead to its ejection into interstellar space, where encounters with stars can continue the process.

So, regardless of where you want to go, "escape velocity" in its common meaning suffices. However, in general, low initial velocity trajectories take a long time. Investigators seeking results in their lifetimes may want to choose trajectories with higher initial velocities for their spacecraft.

$\endgroup$
0
$\begingroup$

enter image description here

to obtain the escape velocity of a mass $~m~$ unter the gravitation forces of the sun and the earth , we first transformed , the velocities to the center of mass velocity and the relative velocities . with

$$ \dot r_{{C}}=0={\frac {M_{{S}}\dot r_{{S}}+M_{{E}}\dot r_{{E}}+m\dot r_{{m}}}{M_{{S}}+M_{{E} }+m}} \\\dot r_{ES}=\dot r_E-\dot r_S\\ \dot r_{mS}=\dot r_m-\dot r_S$$

solve those 3 equations for $~\dot r_S=v_S~,\dot r_E=v_E~,\dot r_m=v_m~$

$$v_S=-{\frac {M_{{E}}v_{{{\it ES}}}}{M_{{S}}+M_{{E}}+m}}-{\frac {mv_{{{\it mS}}}}{M_{{S}}+M_{{E}}+m}} $$

$$v_E={\frac { \left( M_{{S}}+m \right) v_{{{\it ES}}}}{M_{{S}}+M_{{E}}+m}}- {\frac {mv_{{{\it mS}}}}{M_{{S}}+M_{{E}}+m}}$$

$$v_m=-{\frac {M_{{E}}v_{{{\it ES}}}}{M_{{S}}+M_{{E}}+m}}+{\frac { \left( M_ {{S}}+M_{{E}} \right) v_{{{\it mS}}}}{M_{{S}}+M_{{E}}+m}} $$

with $~r_{mS}\approx r_{ES}=r\quad\Rightarrow v_{mS}\approx v_{ES} =v$

thus the energy is

$$ \frac 12 \frac {v^2\,M_S\,(M_E+m)}{M_S+M_E+m}-\frac {G\,M_S\,M_E}{r}-\frac{G\,M_s\,m}{r}-\frac{G\,M_E\,m}{r_E}=0$$ form here the escape velocity $$v^2\approx 2\,\frac{G\,(M_S+M_E+m)}{r}= 2\,\frac{G\,(M_S+M_E)}{r} $$


$$M_E= 6.370\,10^6 ~,M_S=1.989\,10^{30}~,r=1.496\,10^{11}~,G=6.67430\,10^{-11}$$

thus the escape velocity $~v=42.12~$[km/s]

  • $~r_C~$ center of mass position
  • $~r_S~$ sun position vector inertial system
  • $~r_E~$ earth position vector inertial system
  • $~r_m~$ mass position vector inertial system
  • $~M_S~$ sun mass
  • $~M_E~$ earth mass
  • $~r~$ distance earth sun
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.