Could we launch a missile from a planet with the mass of Jupiter?
Let's assume you mean that Earth now has the mass of Jupiter (as opposed to actually launching from the literal planet Jupiter - whole different question...). Then:
- radius of Earth = $6.4 \times 10^6~\text{m}$
- mass of Jupiter = $1.9 \times 10^{27}~\text{kg}$
- Escape velocity, $v_\text{escape} = \sqrt{\frac{2GM}{r}}$
This gives a value for $v_\text{escape}$ of $200~\text{ km}/\text{s}$. For comparison, the actual value (for the real Earth) is $11~\text{ km}/\text{s}$. Incidentally, the surface gravity on this new Earth is about $300~\text{g}$.
To work out how this could be done, we need the rocket equation, which is $\Delta v = v_\text{exhaust} \ln \frac {m_0} {m_1}$.
We need a delta-V of $200 ~\text{km}/\text{s}\; ,$ with a chemical rocket (exhaust velocity about $4400 ~\text{m}/\text{s}$). Solving for $m_1 = 1~\text{kg}\;,$ we get mass of fuel ($m_0$) required of about $5 \times 10^{16}$ tonnes.
That's about $5\%$ of the mass of all the oceans on Earth. If you used hydrogen and oxygen as the fuel, you'd need to convert a volume equivalent to the Mediterranean Sea.
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 $\Delta v$.
Actually you can go to the orbit of Jupiter with a $\approx 2500$ tonnes rocket and a $3$ tonnes payload. From there you can use an ionic engine. A rocket launched from the equator of Jupiter that turns at $12.6~\text{km}/\text{s}$ needs just an increase in speed $v = 29.5~\text{km}/{\text{s}}$.
$$v_{rj}:= 12.6~{\text{km}}/{\text{s}} \;\;\; R_j := 71492~\text{km}\;\;\; g_j:= 24.79~\text{m}/{\text s^2}$$ Given $$\frac{(v + v_{rj})^2}{R_j}= g_j \\ \text{Find}(v)\rightarrow \left[\frac{1}{5}\cdot \frac{\left[-63~\text{km} + \sqrt{44307167}\cdot (\text{km}\cdot\text{m})^\frac{1}{2}\right]}{s} \; \frac{1}{5}\cdot \frac{\left[-63~\text{km} - \sqrt{44307167}\cdot (\text{km}\cdot\text{m})^\frac{1}{2}\right]}{s}\right] \\= \left(2.95 \times 10^4 \;\; -5.47\times 10^4\right)~\text{m}/\text{s} \\ m_L:= 3000~\text{kg}\;\;\; v_e := 4400~\text{m}/\text{s} \;\;\; v:= 2.95 \times 10^4 ~\text{m}/\text{s}$$ Given $$v=v_e\cdot \ln \left(\frac{m_r}{m_l}\right)\\ \text{Find}(m_r)\rightarrow 2448320.9528130687939~\text{kg}= \; 2.448 \times 10^6~\text{kg}$$