4 ms·

> then you need ~3x the delta-v to get to Mercury compared to the moon. [..] > In theory, you could just use a rocket that's 3x as powerful as the Saturn V to g

> then you need ~3x the delta-v to get to Mercury compared to the moon. [..] > In theory, you could just use a rocket that's 3x as powerful as the Saturn V to get enough fuel into orbit to get to Mercury via a direct route I'm a bit confused. I have a vague memory that due to the Tsiolkovsky rocket equation the amount of fuel needed to reach a delta-v grow exponentially. How is it possible that you need a 3x bigger rocket to reach 3x the delta-v?

sxp 5y ago
I'm fuzzing the math a bit, but assume you have a Saturn V style rocket that can get 100 tons into orbit. And that consists of 50 tons of spaceship and 50 tons of fuel. Then you get something like V_e * ln((50+50)/50) = .69 * V_e of delta-v. But if you had two more rockets that launched an additional 2 * 100 tons of fuel into orbit, you get V_e * ln((250+50)/50) = 1.79 * V_e. This gives you 2.5x the delta-v. Decrease the propellant mass fraction and this value goes up. The numbers are completely made up and based on my time playing KSP (https://xkcd.com/1356/ https://xkcd.com/1356/) rather than real rockets so change the numbers as you see fit. But my core point is that orbital refueling greatly extends the usable delta-v of a rocket once it's prepared in orbit since it's easier to build 3x of a normal rocket than a rocket with 3x the payload.
croutonwagon 5y ago
………
- thaumasiotes 5y ago
  > When you escape earth orbit you have to exceed its orbital velocity around the sun. This doesn't sound right. Orbital and escape velocities for the earth should depend only on the mass of the earth.
  - bagels 5y ago
    You can escape earth's orbit going barely faster, or barely slower than earth relative to the sun, depending on where you're going. This stuff can be confusing, because you're mixing two different reference frames (earth and the sun).
    - hoseja 5y ago
      That sounds like you're assuming you're standing still beforehand. Sure, the effect afterwards might look like you're going around the sun faster/slower than Earth but you started deep in Earth's gravitational well.
    - exDM69 5y ago
      This analysis falls short if you apply three body orbital mechanics. Going barely faster or slower than the Earth in close proximity to Earth will slowly cause Earth's gravity to pull the spacecraft back in. This can mean all sorts of funny shaped trajectories (horseshoe asteroids etc). You need about 11-12 km/s to relative to the Earth to escape Earth's gravitational pull from LEO (only a bit more than it takes to reach the moon). Earth itself is orbiting the sun at about 30 km/s. To not get re-captured, the spacecraft needs to be far enough in distance or the difference in velocity needs to be large enough. How much is "enough" depends on the time scale, but for spacecraft we're talking several km/s.
thescriptkiddie 5y ago
One big factor that the Tsiolkovsky rocket equation fails to account for is the square–cube law. When you scale up a rocket, the mass of the structure of the rocket tends to grow more slowly than the mass of the fuel it contains, which more or less cancels out the log term in the rocket equation. Note that this does not necessarily hold true for extremely small or extremely large rockets.
- ithkuil 5y ago
  But even if the mass of the rocket structure was zero, you still need fuel to move the fuel right?

sxp 5y ago

I'm fuzzing the math a bit, but assume you have a Saturn V style rocket that can get 100 tons into orbit. And that consists of 50 tons of spaceship and 50 tons of fuel. Then you get something like V_e * ln((50+50)/50) = .69 * V_e of delta-v. But if you had two more rockets that launched an additional 2 * 100 tons of fuel into orbit, you get V_e * ln((250+50)/50) = 1.79 * V_e. This gives you 2.5x the delta-v. Decrease the propellant mass fraction and this value goes up. The numbers are completely made up and based on my time playing KSP (https://xkcd.com/1356/ https://xkcd.com/1356/) rather than real rockets so change the numbers as you see fit. But my core point is that orbital refueling greatly extends the usable delta-v of a rocket once it's prepared in orbit since it's easier to build 3x of a normal rocket than a rocket with 3x the payload.

croutonwagon 5y ago

………

thaumasiotes 5y ago
> When you escape earth orbit you have to exceed its orbital velocity around the sun. This doesn't sound right. Orbital and escape velocities for the earth should depend only on the mass of the earth.
- bagels 5y ago
  You can escape earth's orbit going barely faster, or barely slower than earth relative to the sun, depending on where you're going. This stuff can be confusing, because you're mixing two different reference frames (earth and the sun).
  - hoseja 5y ago
    That sounds like you're assuming you're standing still beforehand. Sure, the effect afterwards might look like you're going around the sun faster/slower than Earth but you started deep in Earth's gravitational well.
  - exDM69 5y ago
    This analysis falls short if you apply three body orbital mechanics. Going barely faster or slower than the Earth in close proximity to Earth will slowly cause Earth's gravity to pull the spacecraft back in. This can mean all sorts of funny shaped trajectories (horseshoe asteroids etc). You need about 11-12 km/s to relative to the Earth to escape Earth's gravitational pull from LEO (only a bit more than it takes to reach the moon). Earth itself is orbiting the sun at about 30 km/s. To not get re-captured, the spacecraft needs to be far enough in distance or the difference in velocity needs to be large enough. How much is "enough" depends on the time scale, but for spacecraft we're talking several km/s.

thescriptkiddie 5y ago

One big factor that the Tsiolkovsky rocket equation fails to account for is the square–cube law. When you scale up a rocket, the mass of the structure of the rocket tends to grow more slowly than the mass of the fuel it contains, which more or less cancels out the log term in the rocket equation. Note that this does not necessarily hold true for extremely small or extremely large rockets.

ithkuil 5y ago
But even if the mass of the rocket structure was zero, you still need fuel to move the fuel right?