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

Question

Suppose you have a space rocket which, launching from Earth, has a velocity relative to Earth of 11 km/s after escaping Earth. From what I've read in another topic in Space SE, the orbital velocity of the planet where a space rocket launches plays a role in how long the travel to another solar system object will take. So on the one hand Earth's orbital velocity being faster than Ceres' would play a role in making a space rocket launching from Earth faster relative to the sun than one launching from Ceres I suppose, but on the 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 relative to the sun than one launching from Earth? Is it possible to calculate its speed relative to the sun when launching from Ceres, for the same rocket with a speed capacity of 11 km/s relative to Earth when launching from Earth?

Answer 1

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!

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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.

Answer 2

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?

Given that you are talking about achieving the maximum velocity relative to the Sun for rockets with the same capability, you are not meaning a Hohmann transfer (where the required $\Delta v$ would depend on the body being launched from). Therefore, the destination is not relevant other than it being further out in the solar system (requiring an increase in Sun relative velocity to reach, rather than the decrease required for reaching targets closer to the Sun).

I will therefore assume the velocity at escape is aligned with the solar orbit of the body being launched from, maximizing the resulting velocity relative to the Sun. Such an orbit (with sufficient velocity) will intersect the orbit of Jupiter (ignoring the inclination difference, which is only 1.3° between Earth and Jupiter). If timed correctly, Jupiter will be at the intersect point when the intersect happens, so it is a valid trajectory for reaching Jupiter.

Suppose you have a space rocket which when launching from Earth its speed relatively to Earth it's 11 km/s.

Based on your comments on uhoh's answer, you mean that the velocity after escaping Earth is 11 km/s relative to Earth. The velocity at a 200 km perigee (a reasonable height for an Earth escape trajectory) can then be determined by the formula $v^2 = v_e^2 + v_\infty^2$. In this case $v_\infty$ is 11 km/s.

To calculate $v_e$, we can use the escape velocity formula: $v_e = \sqrt {\frac {2GM} {d}}$, where GM is the standard gravitational parameter and d is the distance from the object. Earth has a standard gravitational parameter of $3.986\times10^{14} m^3/s^2$. To determine the distance, we can add the radius of the Earth (6371 km) to the height above the surface (200 km), giving a result of 6571 km. Substituting in those values gives an escape velocity of 11,015 m/s.

Substituting in the escape velocity, we get a perigee velocity of 15,567 m/s. I will ignore any losses for reaching 200 km, and therefore assume the rocket has a $\Delta v$ of 15,567 m/s.

We can then determine the velocity relative to Ceres after escaping Ceres. The same formula ($v^2 = v_e^2 + v_\infty^2$) can be used, but this time we have v, and need to calculate $v_e$, to determine $v_\infty$. Reorganizing the equation to be in terms of $v_\infty$, we obtain $v_\infty = \sqrt {v^2 - v_e^2}$

Ceres has a standard gravitational parameter of $6.263\times10^{10} m^3/s^2$, and I will assume a starting altitude of 50 km. Ceres has a radius of 469.7 km, which gives a total distance of 519.7 km. Substituting in those values, we get an escape velocity of 491 m/s.

Substituting in the rocket's $\Delta v$ and the escape velocity, we get a $v_\infty$ from Ceres of 15,559 m/s. The weak gravity of Ceres has almost no effect on the rocket's velocity, only reducing it by 8 m/s.

The next step is to determine the velocity relative to the Sun. Earth has a maximum orbital velocity of 30.29 km/s. Adding the 11 km/s $v_\infty$ velocity, the spacecraft's velocity relative to the Sun is 41.29 km/s. Ceres has an average orbital velocity of 17.92 km/s. Adding the 15,559 m/s $v_\infty$, the spacecraft's Sun relative velocity is 33,479 m/s. This value is lower than for the spacecraft launched from Earth, but comparing these values is not very meaningful since they are at different heights from the Sun.

To compare these values more directly, we can calculate a solar $v_\infty$ value for both spacecraft. The Sun has a standard gravitational parameter of $1.3271\times10^{20} m^3/s^2$, while the Earth at perihelion is 147.1 million km away from the Sun (escaping at perihelion is optimal due to the Oberth effect). This gives a solar escape velocity of 42,478 m/s, which is faster than the spacecraft. Therefore the spacecraft launched from Earth will not escape the Sun, so a $v_\infty$ value is meaningless (the value would be imaginary). Ceres has a 414 million km semi-major axis, giving a solar escape velocity of 25,320 m/s. This means that, unlike the Earth-launched spacecraft, the spacecraft launched from Ceres will escape the Sun (and the precise $v_\infty$ value is unnecessary to determine which spacecraft is more capable).

In conclusion, while the initial Sun relative velocity of the Earth-launched spacecraft is higher than the Ceres-launched spacecraft, the Ceres-launched spacecraft is capable of escaping the Sun, while the Earth-launched spacecraft is not. This means that launching from Ceres provides increased capability compared to launching from Earth.