I remember reading here that it is more difficult (cost? fuel?) to send a satellite towards the Sun compared to sending a satellite away from it. Is this true? And if it is, how come?
$\begingroup$
$\endgroup$
0
Add a comment
|
2 Answers
$\begingroup$
$\endgroup$
1
The ratio of orbital velocities of Mercury and Earth is about $$(47.87 km/s) / (29.78 km/s) = 1.607.$$ Potential energy (negative) plus orbital kinetic energy is propotional to the square of the ratio of the orbital velocities, hence to $$1.607^2 = 2.58$$ for the Earth - Mercury example.
So you need more energy (the 1.58-fold) to slow down from Earth to Mercury relative to the sun than to escape from Earth orbit out of the solar system.
-
$\begingroup$ More simply, the ratio of the specific energy of Mercury and Earth is one over the ratio of their semi-major axes from the Sun. So $2.58={1\,\mathrm{au}\,/\,0.387\,\mathrm{au}}$. $\endgroup$ Commented Feb 6, 2014 at 6:36
$\begingroup$
$\endgroup$
8
Well, the way you said it isn't quite right. You could spend the same $\Delta V$ to escape Earth in one direction as the other, and in one direction you would move away from the Sun, and in the other direction you would move towards it.
What I think you're referring to is the fact that it would take far less $\Delta V$ from Earth to escape the Solar System than it would to reach the Sun and touch its surface. From low-Earth orbit, it takes $8.75\,\mathrm{km/s}$ to escape the Solar System completely, but it takes $16.8\,\mathrm{km/s}$ to dive into the Sun from low-Earth orbit.
Of course, you wouldn't do that $\Delta V$ yourself. In either direction, you would use Jupiter to send you in or out (or perhaps even in for the out case, if you want to do an escape maneuver near the Sun, which can be very efficient).
answered Feb 6, 2014 at 3:06
-
$\begingroup$ It takes 8.75 km/s to get from the surface of Earth to the lower Earth orbit. For the escape velocities I'm getting higher $\Delta v$. $\Delta v$ translates to kinetic energy via $E=0.5 mv^2$. Crushing into the sun takes about the same $\Delta v$ as leaving the solar system. Adjusting to a reasonable orbit around the sun, e.g. to a Mercury orbit needs additional deceleration. The slingshot way is agreed. (See also escape velocities: en.wikipedia.org/wiki/Escape_velocity) $\endgroup$– GeraldCommented Feb 6, 2014 at 6:10
-
2$\begingroup$ If you like, you can post a question with details on your attempt at the calculation to find out where you went wrong. $\endgroup$ Commented Feb 6, 2014 at 6:30
-
$\begingroup$ At the moment I've not the time to discuss this in detail. May be later this week. There are many ways to design trajectories. I've been referring to the direct way outside the field of gravity of the planets without exploiting flyby techniques. $\endgroup$– GeraldCommented Feb 6, 2014 at 6:49
-
$\begingroup$ Might it be, your numbers are miles/second? I get a similar number for the escape from the solar system, when dividing my result by 1.7. $\endgroup$– GeraldCommented Feb 6, 2014 at 11:38
-
$\begingroup$ It's a factor of 1.772 between your and my numbers for a direct dive into the sun from Earth - Sun distance outside Earth's field of gravity on a geosynchronous orbit. $\endgroup$– GeraldCommented Feb 6, 2014 at 11:46