# Energy Conservation: Minimum Launch Speed

You are on Jupiter and would like to send a probe into space so that it does not fall back to the surface. What minimum launch speed do you need if Jupiter's mass is 1.9 x 10^27?

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## SOLUTION This solution is **FREE** courtesy of BrainMass!

Note that in order to calculate the escape velocity one must know the radius of the planet.

The idea here is that the Jupiter's gravitational potential energy at infinity is zero.

If at infinity there will be a non-zero potential energy - there will be a force exerted by Jupiter on the probe and if the spaceship stops moving at infinity, it will start falling back to Jupiter.

Therefore, the minimal condition for the velocity, that at infinity both potential energy and kinetic energy are zero.

Energy is conserved, so this must be the condition at the initial point - total mechanical energy equals zero.

On Jupiter, the potential energy is:

U = -GMm/r

Where G is the gravitational constant, M is Jupiter's mass, m is the mass of the probe and r is Jupiter's radius.

The kinetic energy at blast-off is simply: K = 1/2*m*v^2

Where v is the initial speed.

The total mechanical energy is:

E = U+K = -GMm/r + (mv^2)/2

This value must be equal to the total energy at infinity, namely - zero.

Hence:

-GMm/r + (mv^2)/2

mv^2 = 2GMm/r

v= sqrt(2GM/r) sqrt means square root.

All that is left is to plug in the numbers:

M = 1.9 x 10^27 kg

G = 6.67 x 10^(-11) m^3/(kg x s ^2)

r = 71492 km = 71.5 x 10^6 m

v = sqrt(2GM/r) = sqrt (3.5 x 10^9) = 5.9 x 10^4 m/s

So the escape velocity from Jupiter is about 59 km/s.

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