# Would a space rocket launching from Ceres in a path to Jupiter be faster relatively to the sun than one launching from Earth?

Suppose you have a space rocket which when launching from Earth its speed relatively to Earth it's 11 km/s. For what I've read in another topic in Space SE, the orbital velocity of the planet where a space rocket launchs plays a role in how long the travel to another solar system object will take. So in one hand Earth's orbital velocity faster than Ceres' would play a role to make a space rocket launching from Earth faster relatively to the sun than one launching from Ceres I suppose, but in other hand, Ceres 0.029 g gravity should make that space rocket launching from Ceres faster I suppose, since the force of the space rocket thrust would be counteracted less by the lower gravity force of the planet.

Would a space rocket launching from a low gravity Ceres in a path to Jupiter, be faster or slower relatively to the sun than one launching from Earth? Is it possible to calculate its speed relatively to the sun when launching from Ceres, for the same rocket with a speed capacity of 11 km/s relatively to Earth when launching from Earth?

On the down side, at 2.77 times the distance from the Sun as Earth, the orbital speed of Ceres is $$\sqrt{1 / 2.77} = 0.6$$ of Earth's orbital speed.

But on the up side, it's also starting higher up in the Sun's gravitational potential so it needs less delta-v to get to Jupiter, and as you point out, Ceres has a far lower escape velocity, which also helps!

Ignoring atmosphere, here's *a simplistic attempt( to calculate the general case of going from a planet with mass $$M$$ and radius $$R$$ orbiting the Sun at semimajor axis $$a_1$$ to an elliptical orbit with aphelion of the other planet's semimajor axis $$a_2$$.

To escape first planet's gravity:

$$v_{escape} = \sqrt{\frac{2GM}{R}}$$

Current heliocentric velocity after escape:

$$v_1 = \sqrt{\frac{GM_{Sun}}{a_1}}$$

Necessary heliocentric velocity to achieve aphelion $$a_2$$ using vis-viva equation with $$r=a_1$$ and $$a=(a_1+a_2)/2$$:

$$v = \sqrt{GM_{Sun} \left( \frac{2}{a_1} - \frac{1}{(a_1+a_2)/2} \right)} = \sqrt{2 GM_{Sun} \left( \frac{1}{a_1} - \frac{1}{a_1+a_2} \right)}$$

Total delta-v needed

$$\Delta v = v_{escape} + (v - v_1)$$

$$\Delta v = \sqrt{\frac{2GM}{R}} + \sqrt{2 GM_{Sun} \left( \frac{1}{a_1} - \frac{1}{a_1+a_2} \right)} - \sqrt{\frac{GM_{Sun}}{a_1}}$$

From Earth, using a standard gravitational parameter $$GM$$ of 3.986E+14 m^3/s^2 and radius $$R$$ of 63787137 meters, we get an escape velocity of 11,200 m/s as you point out. With $$a_1$$ of 1.5E+11 meters and Jupiter's $$a_2 = 5.2 a_1$$ the total delta-v for an aphelion at Jupiter is 20,000 m/s, which is out of the range of our 11 km/s rocket.

From Ceres, using a standard gravitational parameter $$GM$$ of only 6.263+10 m^3/s^2 and radius $$R$$ of only 470,000 meters, we get an escape velocity of only 500 m/s which is far lower than Earth's as you expected. With $$a_1$$ of 1.5E+11 meters and Jupiter's $$a_2 = 5.2 a_1$$ the total delta-v for an aphelion at Jupiter staring from Ceres is only 3000 m/s, which is way smaller than our 11 km/s rocket can supply. So the mission is a success!

It only took about 2,500 m/s to get an aphelion at Jupiter's 5.2 AU from Ceres' 2.77 AU. There is plenty of delta-v left to circularize at Jupiter's 5.2 AU and drop into a very high orbit around it, but you'll need more delta-v than you have to get into a low orbit near one of the Galilean moons From your circularized heliocentric orbit matching Jupiter, you'd need about 19,000 m/s to drop down to Europa's tiny 670,000 km orbit for example.

• So if I understood correctly, you calculated that from Earth you can't reach Jupiter, but from Ceres you can. But I also liked to know (if possible) if the same space rocket which was designed to go 11 km/s relatively to Earth to escape Earth , if it could reach a faster speed if launched from Ceres, refering its speed relative to the sun (since Ceres has lower gravity but also lower orbital speed). In this graph the speed of Insight is Earth orbit speed + speed of Insight leaving Earth, right? en.wikipedia.org/wiki/Hohmann_transfer_orbit#/media/… – Pablo Feb 15 at 2:24
• @Pablo oh, I see what you mean. Yes I think that's correct. Okay in a few hours I will modify my answer and add more. Thanks for the quick comment! – uhoh Feb 15 at 2:33
• You're overestimating the delta-V. Adding escape velocity to the required Heliocentric velocity only matches the delta-V spent if your flight plan is "burn to exactly escape velocity, coast to interplanetary space, and do a second burn in interplanetary space to perform the transfer." This answer appears to walk through the right calculations for single-burn at orbit and destination. space.stackexchange.com/questions/1380/… – notovny Feb 15 at 12:06
• @notovny That sounds interesting. I know that the whole Oberth effect thing tells us that it matters very much where in a potential field one burns. Okay I'll "do the math" properly and rewrite. Thanks! – uhoh Feb 16 at 0:30
• Hi @Pablo better not accept my answer yet! I need to make some edits since I need to address your comment and (at)notovny's as well. I'm having a busy weekend+Monday but I'll get to it in a day or so. Thanks! – uhoh Feb 17 at 4:00