Question

Figure 13-53 is a graph of the kinetic energy Kof an asteroid versus its distance rfrom Earth’s center, as the asteroid falls directly in toward that center. (a) What is the (approximate) mass of the asteroid? (b) What is its speed at$\mathbf{r}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{945}\mathbf{\times }{\mathbf{10}}^{\mathbf{7}}\mathbf{m}$ ?

Short answer

Answer:

1. The mass of asteroid is$m=1.0\times {10}^{3}kg$
2. The speed of asteroid is at $r=1.945\times {10}^{7}misv=1.5\times {10}^3 \text{m/s}$

Step-by-step solution

Significance of Newton’s law of universal gravitation

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

We can use the concept of energy conservation law. When the asteroid falls towards the earth, its gravitational potential energy is converted into kinetic energy.

Formula:

$$\begin{gathered} U=-\frac{GMm}{r} \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

Where.

G is the gravitational constant ( $6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$)

M is the mass earth

K is the kinetic energy

U is the potential energy

v is the speed of object

m is the mass of asteroid

(a) Determining the mass of asteroid

According to the energy conservation law,

$$\begin{gathered} K_1-U_1=K_2-U_2 \\ {K}_1-\frac{GMm}{r_1}=K_2-\frac{GMm}{r_2} \\ \frac{GMm}{r_2}-\frac{GMm}{r_1}=K_2-K_1 \\ m=\frac{K_2-K_1}{GM\left(\frac{1}{r_2}-\frac{1}{r_1}\right)} \\ =\frac{3.2\times {10}^9\text{J}-2.2\times {10}^9\text{J}}{\left(6.67\times {10}^{-11}\frac{{\text{N.m}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\left(5.98\times {10}^{24}\text{kg}\right)\left(\frac{1}{1.77\times {10}^7\text{m}}-\frac{1}{1.85\times {10}^7\text{m}}\right)} \\ =1.0\times {10}^3\text{kg} \end{gathered}$$

(b) Determining the speed of asteroid

At a distance $r=1.945\times {10}^{7}m$, the kinetic energy of the asteroid is $K=1.2\times {10}^{9}J$. Hence the speed of the asteroid is

$K=\frac{1}{2}m{v}^2$

$$\begin{gathered} dv=\sqrt{\frac{2K}{m}} \\ =\sqrt{\frac{2\times 1.2\times {10}^{9}J}{1.0\times {10}^{3}kg}} \\ =1.5\times {10}^{3}m/s \end{gathered}$$

The speed of asteroid is at r = 1.945 x 10⁷ is v =$1.5\times {10}^{3}m/s$