Chapter 8, Problem 65P

To determine

To plot: Acceleration a against probe position $\frac{1}{r^2}$.

To determine: The mass of the asteroid.

Answer to Problem 65P

The mass of the asteroid is $\underset{\_}{1.58\times {10}^{16}\text{kg}}$.

Explanation of Solution

Given information:

The acceleration and position of probe data are given.

Calculation:

The gravitational acceleration at the surface of the asteroid is shown below:

$a=\frac{GM}{r^2}$

Here, G is the constant of universal gravitation with the value of $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$, $M_{\text{S}}$ is the mass of the asteroid, and $r$ is the position of probe.

Find the value of $\frac{1}{r^2}$ and plot a vs $\frac{1}{r^2}$ to obtain the best fit line.

The slope of the graph provides the value of GM.

Divide the slope value by G to obtain the mass of the asteroid.

Convert the acceleration value of $\text{0}\text{.172}{\text{mm/s}}^{\text{2}}$ to ${\text{m/s}}^{\text{2}}$:

$\begin{array}{c}a\text{ = 0}\text{.172}{\text{mm/s}}^{\text{2}}\\ \text{= 0}\text{.172}{\text{mm/s}}^{\text{2}}\left(\frac{{10}^{-3}\text{m}}{1\text{mm}}\right)\\ =\text{0}\text{.000172}{\text{m/s}}^{\text{2}}\end{array}$

Find the value of $\frac{1}{r^2}$ for the r value of 80.0 km.

$\begin{array}{c}\frac{1}{r^2}=\frac{1}{{\left(80.0\text{km}\right)}^2}\\ =\frac{1}{{\left(80.0\text{km}\times \frac{{10}^3\text{m}}{1\text{km}}\right)}^2}\\ =1.5625\times {10}^{-10}\end{array}$

Similarly convert the remaining acceleration values and also find the value of $\frac{1}{r^2}$ for the remaining data.

Table 1 shows the acceleration (a) and $\frac{1}{r^2}$ values .

Acceleration $a\left({\text{mm/s}}^{\text{2}}\right)$	Acceleration $a\left({\text{m/s}}^{\text{2}}\right)$	Position of the probe $r\left(\text{km}\right)$	Value of $\frac{1}{r^2}$
0.172	0.000172	80.0	$1.5625\times {10}^{-10}$
0.353	0.000353	55.0	$3.3058\times {10}^{-10}$
0.704	0.000704	40.0	$6.25\times {10}^{-10}$
0.858	0.000858	35.0	$8.1633\times {10}^{-10}$
1.18	0.00118	30.0	$1.1111\times {10}^{-9}$

Use Column (2) and Column (4) values to plot $a\left({\text{m/s}}^{\text{2}}\right)$ versus $\frac{1}{r^2}$.

Sketch of $a\left({\text{m/s}}^{\text{2}}\right)$ versus $\frac{1}{r^2}$ is shown in Figure 1.

Refer Figure 1.

The slope of the line is $=1.054\times {10}^6$.

Slope is equal to GM.

$GM=1.054\times {10}^6$ (1)

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for G to obtain the mass of the asteroid.

$\begin{array}{c}\left(6.67\times {10}^{-11}\right)M=1.054\times {10}^6\\ M=\frac{1.054\times {10}^6}{6.67\times {10}^{-11}}\\ =1.58\times {10}^{16}\text{kg}\end{array}$

Therefore, the mass of the asteroid is $\underset{\_}{1.58\times {10}^{16}\text{kg}}$.