Question

The gravitational attraction between two objects with masses $m_1$ and $m_2$, separated by distance x, is $F=G{m}_1{m}_2/x^2$ , where G is the gravitational constant.

a. How much work is done by gravity when the separation changes from $x_1$ to $x_2$? Assume $x_2<x_1$.

b. If one mass is much greater than the other, the larger mass stays essentially at rest while the smaller mass moves toward it. Suppose a $1.5\times {10}^{13}kg$ comet is passing the orbit of Mars, heading straight for the sun at a speed of $3.5\times {10}^{4}m/s$ What will its speed be when it crosses the orbit of Mercury? Astronomical data are given in the tables at the back of the book, and $G=6.67\times {10}^{-11}N{m}^2/k{g}^2.$ .

Short answer

a. The required amount of work done by gravity is $G{m}_1{m}_2\left(\frac{x_2-x_1}{x_2{x}_1}\right)$

b. The speed of comet when it passes mercury's orbit is$2.1\times {10}^{5}m/s$

Step-by-step solution

Content Introduction

The work done by force F on a particle if particle moves in the same direction as that of the force is

$W=-{\int }_{x_i}^{x_f}Fdx$

Negative sign represents that the force is towards the center.

The expression for work energy theorem is

$W=K{E}_f-K{E}_i$

Here, $K{E}_f$is the final kinetic energy of the comet and$K{E}_i$is the initial kinetic energy of the comet.

Explanation (Part a)

Substitute $\frac{G{m}_1{m}_2}{x^2}$for F, $x_{1}for{x}_{i,}{x}_{2}for{x}_f$

$$\begin{gathered} W=-{\int }_{x_1}^{x_2}\frac{G{m}_1{m}_2}{x^2}dx \\ W=-G{m}_1{m}_2{\int }_{x_1}^{x_2}\frac{1}{x^2}dx \\ W=-G{m}_1{m}_2[\frac{1}{x_2}-\frac{1}{x_1}] \\ W=G{m}_1{m}_2\left[\frac{x_1-x_2}{x_2{x}_1}\right] \end{gathered}$$

Explanation (Part b)

Apply the work energy theorem to calculate final velocity of comet.

$$\begin{gathered} W=K{E}_f-K{E}_i \\ W=\frac{1}{2}{m}_1({v_f}^2-{v_i}^2) \end{gathered}$$

Substitute$G{m}_1{m}_2\left[\frac{x_1-x_2}{x_2{x}_1}\right]forW.$

localid="1647795652613" $G{m}_1{m}_2\left[\frac{x_1-x_2}{x_2{x}_1}\right]=\frac{1}{2}{m}_1({v_f}^2-{v_i}^2)$

Rearrange above expression for the final velocity of comet.

$$\begin{gathered} G{m}_1{m}_2\left[\frac{x_1-x_2}{x_2{x}_1}\right]=\frac{1}{2}{m}_1({v_f}^2-{v_i}^2) \\ \frac{1}{2}{m}_1{v_f}^2=G{m}_1{m}_2\left[\frac{x_1-x_2}{x_2{x}_1}\right]+\frac{1}{2}{m}_1{v_i}^2 \\ {v_f}^2=2G{m}_2\left[\frac{x_1-x_2}{x_2{x}_1}\right]+{v_i}^2 \end{gathered}$$

Substitute values and evaluate.

role="math" localid="1647796219724" $$\begin{gathered} v_f=[\left[\begin{array}{c}(6.67\times {10}^{-11}N.m^2/k{g}^2)\\ (1.99\times {10}^{30}kg)\end{array}\right]\left[\begin{array}{c}(2.28\times {10}^{11}m)-(5.79\times {10}^{9}m)\\ (2.28\times {10}^{11}m)(5.79\times {10}^{9}m\end{array}\right]+{(3.5\times {10}^{4}m/s)}^2 \\ {v}_f=2.1\times {10}^{5}m/s \end{gathered}$$