Question

problem 6:
a visitor to mars is ready to leave in her rocketship. find the escape velocity the rocketship needs to escape mars, given that the mass of mars ( m = 6.39\times10^{23}\text{ kg} ) and the radius is ( r = 3,390\text{ km} ). make sure to show steps, starting from the conservation of energy.

Explanation:

Step1: Conservation of energy

The total mechanical energy at the surface of Mars \(E_1=\frac{1}{2}mv^2-\frac{GMm}{R}\), and at infinity \(E_2 = 0\). By conservation of energy \(E_1=E_2\).

Step2: Solve for escape velocity \(v\)

From \(\frac{1}{2}mv^2-\frac{GMm}{R}=0\), we can cancel out \(m\) (mass of the rocket - ship). Then \(\frac{1}{2}v^2=\frac{GM}{R}\).
We know that \(G = 6.67\times10^{-11}\space N\cdot m^2/kg^2\), \(M = 6.39\times 10^{23}\space kg\), \(R=3.39\times 10^{6}\space m\).
Substitute the values into the formula \(v=\sqrt{\frac{2GM}{R}}\).

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\(\frac{2\times6.67\times6.39}{3.39}\approx25\), so \(v=\sqrt{25\times 10^{6}} = 5\times10^{3}\space m/s\)

Answer:

The escape velocity of the rocket - ship is \(5\times 10^{3}\space m/s\)