2
$\begingroup$

I'm not scientifically adept, so bear with me. Would it be feasible to retain enough fuel on new satellites/space stations, etc., that we could dispose of them when obsolete by sending them to the sun where they would likely incinerate before they came close to the "surface."

How would we ensure burn up rather orbiting the sun? Already in earth orbit, how much speed would be needed to "hit" the sun?

Impractical because of fuel needs? We could use the space station as an example. It would seem to me that eventually the gravity of the Sun would pull it lessening the need for fuel, it's getting way from the rest of solar system's gravities that would require fuel.

$\endgroup$
8
  • 11
    $\begingroup$ You'd actually need more fuel to send a payload into the Sun from Earth than you would need to send the same payload completely out of the Solar System. astronomy.com/science/… $\endgroup$ Nov 18, 2023 at 1:21
  • 3
    $\begingroup$ Apparently, it is not so easy to deliberately get close to the sun: the Parker Solar Probe uses gravity assists from Venus and it will use 7 of them. And not even get to the sun. $\endgroup$
    – Ed V
    Nov 18, 2023 at 1:21
  • 2
    $\begingroup$ What Solomon said. In fact, the delta-v needed to escape the Solar System is only ~0.4 (to be more precise, $\sqrt2-1$) the delta-v needed to drop the satellite into the Sun. $\endgroup$
    – PM 2Ring
    Nov 18, 2023 at 4:05
  • 2
    $\begingroup$ Would Space Exploration be a better home for this question? $\endgroup$
    – Qmechanic
    Nov 18, 2023 at 10:56
  • 1
    $\begingroup$ @SolomonSlow that's not really applicable to disposal through...why do a slingshot around Jupiter when you could just drop things into it? $\endgroup$ Nov 18, 2023 at 12:58

3 Answers 3

8
$\begingroup$

For an object to fall into the Sun, from close to the Earth, its angular momentum, which is conserved in any ballistic trajectory, would have to be reduced close to zero - since its velocity vector would need to point close to the centre of the Sun.

An object in orbit around the Earth has a substantial velocity component of $\sim 30$ km/s tangential to the Sun, and hence a large angular momentum, simply as a result of the Earth's orbital speed around the Sun.

To reduce this tangential velocity component (close to zero), in order to remove angular momentum and send something sunward, requires huge energy expenditure. As pointed out in comments, it would take less energy to send something out of the Solar System entirely.

$\endgroup$
5
$\begingroup$

The Earth is moving around the Sun at about 30 km/s so all the satellites orbiting the Earth share this velocity (plus or minus their orbital velocity relative to the Earth).

To fall into the Sun this 30 km/s orbital velocity has to be reduced to (approximately) zero, and this would require a large amount of thrust. You would have to equip your satellite with a massive engine and a massive amount of fuel to make this possible.

For more on this see the question Orbital Mechanics and Launching into the Sun from the Space Exploration Stack Exchange.

$\endgroup$
4
$\begingroup$

In addition, to the already decent answers, I'll provide some numbers which will prove the point. Earth's tangential speed in orbit is about $30~km/s$. Let's take a typical satellite mass as $1100~kg$. So, to fully cancel it's orbital speed around the Sun, you would have to supply : $$ K = 1100~kg × 0.5 × (-30~km/s)^2 \approx 500~\text{GJ} \tag 1$$

That's the energy which typical nuclear power plant of $1~\text{GW}$, which powers electricity for the whole cities, produces in the continuous operation for $8~\text{minutes}$. Needles to say, that it's a huge energy amount.

Using Tsiolkovsky rocket equation, we can calculate how much liquid fuel satellite must burn for reaching negative ($-30~km/s$) velocity with respect to tangential orbital velocity given by gravity of sun (so that you would get total tangential velocity of $0~ km/s$) :

$$ m_p = m_s\left( e^{{\Delta v}/{v_e}} - 1\right) \tag 2,$$

where $m_s$ is empty satellite mass (without fuel), $v_e$ is exhaust velocity for liquid propellant, let's take it about $3000~m/s$ for high energy propellants. Substituting values, we get that satellite must burn fully $\approx 24~000 t$ or $24~Ktons$ of liquid fuel.

If we would have to spent such amount of fuel, we would better use it wisely.

$\endgroup$
1
  • 1
    $\begingroup$ For comparison, the Starship Superheavy stack has a combined fuel mass on the order of 4.6 kilotons, or just under 1/5 of that total. 5x that (on orbit no less) to dispose of a 1-ton satellite would be … wasteful, to say the least $\endgroup$ Nov 19, 2023 at 8:18

Your Answer

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