Voyage From The Sun

In this problem, we are looking for the work $W$ needed to move a mass $m$ from a point $R=1.50 \times 10^{11} \text{m}$ from the Sun to a point very far from the Sun. We assume the mass is initially at rest relative to the Sun, and conventionally, when we consider a gravitational escape, we take the escaped object’s final speed to be 0. Therefore, the statement of conservation of energy reduces to $W = \Delta U = U_f - U_0$ since the change in kinetic energy is 0. In other words, the work needed to escape the gravitational influence of the Sun is equal to the object's change in gravitational potential energy.

Furthermore, If we set the zero point of the potential energy at infinity, then $U_f = 0$ and $U_0 = -\frac{G m M_\odot}{R}$ where $M_\odot$ is the mass of the sun, $1.99 \times 10^{30} \text{kg}$ , and $G$ is Newton’s gravitational constant. Plugging in for $U_0$ and $U_f$ in the energy conservation statement above, we find $W = \frac{G m M_\odot}{R}.$ We can divide both sides by mass of the probe, $m$ , to find work per unit mass: $\frac{W}{m} = \frac{G M_\odot}{R}.$

Now we can insert the given parameters in SI units, and upon converting from Joules to tons of TNT, we find $\frac{W}{m} = \frac{(6.67 \times 10^{-11} \frac{\text{J} \cdot \text{m}}{\text{kg}^2})(1.99 \times 10^{30} \text{kg})}{(1.50 \times 10^{11} \text{m})} \times \frac{1 \text{ton of TNT}}{4.18 \times 10^{9} \text{J}} = 0.212 \frac{\text{tons of TNT}}{\text{kg}}.$

Since space probes are several hundreds of kilograms, the energy equivalent of several tens of tons of TNT is needed to boost the probe out of the solar system. In practice, this energy is not provided by propulsion, but rather by strategically using the gravity of orbiting planets.

Recall the minimum speed needed for a body of mass $m$ to escape a gravitational field of a star or planet of mass $M$ from a distance $R$ i.e. the escape velocity is $v_e=\sqrt {2GM/R}$ where G is the Newton's gravitational constant. So, the minimum amount of work needed is simply the minimum kinetic energy you must provide the body to escape. That is $K=\frac {1}{2}mv_e^2 =\frac {1}{2}m\frac {2GM}{R}= \frac {GMm}{R}$ Hence, work per unit mass is $\frac {K}{m}=\frac {GM}{R}$ Plugging in the values and dividing by $4.18×10^9$ J/ton of TNT we get $\boxed{0.2116}$

I found the answer i.e. 0.2124 using another logic. We can integrate the negative of work done per unit mass of probe by moving the probe from near earth to infinity. Plugging the limits and dividing by the TNT Equivalent energy, we get the answer.