Escape Velocity from Mars

To determine the escape velocity of an object of mass \(m\) from a planet, we can use the conservation of mechanical energy since we are ignoring the effects of friction. In this case, the object starts with a speed \(v\) from the surface of Mars of mass \(M\) and radius \(R\); hence, it will initially have a kinetic energy \(K\) and a gravitational potential energy \(U\). The expression for the conservation of the mechanical energy \(E\) reads

\begin{equation}\label{cons}
E_i=E_f,
\end{equation}

where \(i\) and \(f\) are sub-indices used to denote an initial and final state of the body, respectively. Because mechanical energy is the sum of kinetic and gravitational potential energy in this case, we have

\begin{equation}
E=K+U.
\end{equation}

Inserting the result given above into equation\eqref{cons}, we obtain that the conservation of mechanical energy can be written as

\begin{equation}\label{cons2}
K_i+U_i=K_f+U_f.
\end{equation}

The expression for the kinetic energy of a body of mass \(m\) and speed \(v\) is

\begin{equation}\label{kin}
K=\frac{1}{2}mv^2.
\end{equation}

The expression for the gravitational potential energy of a body of mass \(m\) and a body of mass \(M\) that interact through gravity at a distance from centers of mass of \(R\) is

\begin{equation}\label{pot}
U=-\frac{GmM}{R},
\end{equation}

where \(G= 6.67430 \times 10^{-11}\, \text{m}^3\, \text{kg}
^{-1}\,\text{s}^{-2}\) is the Newtonian constant of gravitation. Inserting the expressions given in \eqref{kin} and \eqref{pot} into \eqref{cons2}, we obtain

\begin{equation}\label{cons3}
\frac{1}{2}mv_i^2-\frac{GmM}{R_i}=\frac{1}{2}mv_f^2-\frac{GmM}{R_f}.
\end{equation}

Because the initial distance between centers of mass is the radius of the planet \(R_i=R\) and because we want the object to escape the gravitational attraction, we consider the final state of the body to be infinitely far from the planet; that is \(R_f\to \infty\). As \(R_f\to\infty\) the gravitational potential energy term \(\frac{GmM}{R_f}\to0\); thus, we can write \eqref{cons3} as

\begin{equation}\label{cons4}
\frac{1}{2}mv_i^2-\frac{GmM}{R}=\frac{1}{2}mv_f^2.
\end{equation}

We want to solve for the minimum value of \(v_i\), which occurs in the case where \(v_f=0\), that is, the escape velocity \(v_E\) is the velocity such that at a point infinitely away from the planet, the velocity of the object that escaped is zero. Therefore, equation \eqref{cons4} becomes

\begin{equation}
\frac{1}{2}mv_E^2-\frac{GmM}{R}=0,
\end{equation}

which is equivalent to

\begin{equation}
\frac{1}{2}mv_E^2=\frac{GmM}{R}.
\end{equation}

Canceling out the mass \(m\) on the equation above and multiplying both sides by \(2\), we get

\begin{equation}\label{res}
v_E^2=\frac{2GM}{R}.
\end{equation}

Taking the square-root on both sides of \eqref{res}, we end up with

\begin{equation}
v_E=\sqrt{\frac{2GM}{R}},
\end{equation}

which numerically (using the given values in SI units) is

\begin{equation}
v_E=\sqrt{\frac{2(6.67430 \times 10^{-11}\, \text{m}^3\, \text{kg}
^{-1}\,\text{s}^{-2})(6.39\times10^{23}\,\text{kg})}{3.3895\times10^{6}\,\text{m}}},
\end{equation}

\begin{equation}
v_E\approx 5.016\times 10^{3}\,\text{m/s}.
\end{equation}