We watch two identical astronomical bodies Aand B, each of mass m, fall toward each other from rest because of the gravitational force on each from the other. Their initial center-to-center separation isR_i. Assume that we are in an inertial reference frame that is stationary with respect to the center of mass of this two body system. Use the principle of conservation of mechanical energy (K_f+ U_f=K_i +U_i ) to find the following when the center-to-center separation is 0.5Ri:

(a) the total kinetic energy of the system,

(b) the kinetic energy of each body,

(c) the speed of each body relative to us, and

(d) the speed of body Brelative to body A. Next assume that we are in a reference frame attached to body A(we ride on the body). Now we see body Bfall from rest toward us. From this reference frame, again use$\mathit{K}\mathit{f}\mathbf{}\mathbf{+}\mathbf{}\mathit{U}\mathit{f}\mathbf{}\mathbf{=}\mathbf{}\mathit{K}\mathit{i}\mathbf{}\mathbf{+}\mathit{U}\mathit{i}$to find the following when the center-to-center separation is$\mathbf{0}\mathbf{.}\mathbf{5}\mathit{R}\mathit{i}$:

(e) the kinetic energy of body Band

(f) the speed of body Brelative to body A.

(g) Why are the answers to (d) and (f) different? Which answer is correct?

The mass of each body is m

The initial separation distance is R_i

The final separation distance is 0.5 R_i.

Using the law of conservation of energy, we can find the total K.E of the system. From this, we can find the kinetic energy and velocity of each body relative to the inertial frame of reference or relative to the other body.

Formulae:

$$\begin{gathered} K_i+U_i=K_f+U_f \\ K=\frac{1}{2}m{v}^2 \\ U=-\frac{GMm}{R} \end{gathered}$$

We have the equation for conservation of energy as

$$\begin{gathered} K_i+U_i=K_f+U_f \\ 0-\frac{G{m}^2}{R_i}=K_{total}-\frac{G{m}^2}{0.5R_i} \\ {K}_{total}=\frac{G{m}^2}{R_i} \end{gathered}$$

The total kinetic energy of the system is $K_{total}=\frac{G{m}^2}{R_i}$.

As the mass of each body is the same, their speed should also be the same.

So, the kinetic energy of each body is

$$\begin{gathered} KE=\frac{1}{2}{K}_{total} \\ KE=\frac{G{m}^2}{2R_i} \end{gathered}$$

The kinetic energy of each body is $KE=\frac{G{m}^2}{2R_i}$.

We have the equation for KE as.$K=\frac{1}{2}m{v}^2$

By rearranging this equation for velocity and using the kinetic energy equation from part b, we get

$v=\sqrt{\frac{Gm}{R_i}}$

The speed of each body relative to the inertial frame of reference is $v=\sqrt{\frac{Gm}{R_i}}$.

The relative speed of B with respect to A, is $v_{BA}=2v$.

$v_{BA}=2\sqrt{\frac{Gm}{R_i}}$

The speed of body B relative to body A is $v_{BA}=2\sqrt{\frac{Gm}{R_I}}$.

In this part, the frame of reference is body A, so the total kinetic energy will be equal to the kinetic energy of body B.

Hence

$$\begin{gathered} K_B=K \\ {K}_B=\frac{D{m}^2}{R_i} \end{gathered}$$

Kinetic energy of body B when the frame of reference isbody A is$K_B=\frac{D{m}^2}{R_i}$.

In a similar way to part d,we have

$k_B=\frac{1}{2}m{v^2}_B$

Solving

$v_B=\sqrt{\frac{2Gm}{R_i}}$

The speed of body B relative to body A, measured from frame of reference of A, is$V_B=\sqrt{\frac{2Gm}{R_i}}$

When weconsider bodyA a frame of reference, we neglect the accelerated motion of body A. So, the answer of part f is incorrect, and the answer ofpart d is correct.