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Hey!

The question keeps getting edited! Make up your mind!

You asked about Mars originally, then edited the question. Actual, real Jupiter is flat out impossible. Does it have a surface to launch from? Who knows? What's the pressure at that depth? Can our probes even survive at that depth? Probably not?

What if Earth had the mass of Jupiter? More impossible. It would have a surface gravity of $g \frac{M_J}{M_E}$ or something like 3100 m/s2. I don't think you could even build two-story buildings on that kind of planet

However, here is the math for Mars.

Answer for Mars

Gravity differs, yes, but Mars also has a 0.6 kPa surface pressure, compared to Earth's 100 kPa. This makes comparisons between Earth and Mars practically impossible. Fortunately, the math is easier on Mars.

Tsiolkovsky's rocket equation tells us the answer for general rocket maneuvers.

$\Delta v = v_e \ln \frac{m_0}{m_1}$

For Mars to LMO (low Mars orbit), the $\Delta v$ is about 4.1 km/s. This is just a function of the gravitational potential that you are escaping. For comparison, Earth to LEO is about 9.3-10 km/s, and Kerbin to LKO is about 4.6 km/s.

The value $v_e$ is the effective exhaust velocity, which could be about 4.4 km/s for a bipropellant rocket.

The values $m_0$ and $m_1$ are the masses of the rocket before and after the maneuver.

We will pretend that Mars has no atmosphere.

Suppose that 75% of your rocket is fuel, then $\frac{m_0}{m_1} = \frac{1}{1 - 0.75} = 4$, and your $\Delta v$ is 6.1 km/s, which is more than enough to get into orbit. But it's not enough to escape Mars! For that, you need to double the $\Delta v$.

Gravity differs, yes, but Mars also has a 0.6 kPa surface pressure, compared to Earth's 100 kPa. This makes comparisons between Earth and Mars practically impossible. Fortunately, the math is easier on Mars.

Tsiolkovsky's rocket equation tells us the answer for general rocket maneuvers.

$\Delta v = v_e \ln \frac{m_0}{m_1}$

For Mars to LMO (low Mars orbit), the $\Delta v$ is about 4.1 km/s. This is just a function of the gravitational potential that you are escaping. For comparison, Earth to LEO is about 9.3-10 km/s, and Kerbin to LKO is about 4.6 km/s.

The value $v_e$ is the effective exhaust velocity, which could be about 4.4 km/s for a bipropellant rocket.

The values $m_0$ and $m_1$ are the masses of the rocket before and after the maneuver.

We will pretend that Mars has no atmosphere.

Suppose that 75% of your rocket is fuel, then $\frac{m_0}{m_1} = \frac{1}{1 - 0.75} = 4$, and your $\Delta v$ is 6.1 km/s, which is more than enough to get into orbit. But it's not enough to escape Mars! For that, you need to double the $\Delta v$.

Hey!

The question keeps getting edited! Make up your mind!

You asked about Mars originally, then edited the question. Actual, real Jupiter is flat out impossible. Does it have a surface to launch from? Who knows? What's the pressure at that depth? Can our probes even survive at that depth? Probably not?

What if Earth had the mass of Jupiter? More impossible. It would have a surface gravity of $g \frac{M_J}{M_E}$ or something like 3100 m/s2. I don't think you could even build two-story buildings on that kind of planet

However, here is the math for Mars.

Answer for Mars

Gravity differs, yes, but Mars also has a 0.6 kPa surface pressure, compared to Earth's 100 kPa. This makes comparisons between Earth and Mars practically impossible. Fortunately, the math is easier on Mars.

Tsiolkovsky's rocket equation tells us the answer for general rocket maneuvers.

$\Delta v = v_e \ln \frac{m_0}{m_1}$

For Mars to LMO (low Mars orbit), the $\Delta v$ is about 4.1 km/s. This is just a function of the gravitational potential that you are escaping. For comparison, Earth to LEO is about 9.3-10 km/s, and Kerbin to LKO is about 4.6 km/s.

The value $v_e$ is the effective exhaust velocity, which could be about 4.4 km/s for a bipropellant rocket.

The values $m_0$ and $m_1$ are the masses of the rocket before and after the maneuver.

We will pretend that Mars has no atmosphere.

Suppose that 75% of your rocket is fuel, then $\frac{m_0}{m_1} = \frac{1}{1 - 0.75} = 4$, and your $\Delta v$ is 6.1 km/s, which is more than enough to get into orbit. But it's not enough to escape Mars! For that, you need to double the $\Delta v$.

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Gravity differs, yes, but Mars also has a 0.6 kPa surface pressure, compared to Earth's 100 kPa. This makes comparisons between Earth and Mars practically impossible. Fortunately, the math is easier on Mars.

Tsiolkovsky's rocket equation tells us the answer for general rocket maneuvers.

$\Delta v = v_e \ln \frac{m_0}{m_1}$

For Mars to LMO (low Mars orbit), the $\Delta v$ is about 4.1 km/s. This is just a function of the gravitational potential that you are escaping. For comparison, Earth to LEO is about 9.3-10 km/s, and Kerbin to LKO is about 4.6 km/s.

The value $v_e$ is the effective exhaust velocity, which could be about 4.4 km/s for a bipropellant rocket.

The values $m_0$ and $m_1$ are the masses of the rocket before and after the maneuver.

We will pretend that Mars has no atmosphere.

Suppose that 75% of your rocket is fuel, then $\frac{m_0}{m_1} = \frac{1}{1 - 0.75} = 4$, and your $\Delta v$ is 6.1 km/s, which is more than enough to get into orbit. But it's not enough to escape Mars! For that, you need to double the $\Delta v$.