Chapter 13, Problem X6 We watch two identical astronomical bodies A and B, each

Question

Chapter 13, Problem X6 We watch two identical astronomical bodies A and B, each of mass m = 2.4 x 1027 kg, fall toward each other from rest because of the gravitational force on each from the other. Their initial center-to-center separation is Ri6.6 x 109 m. 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 (Kf UfKiUi) to find the following when the center-to-center separation is 0.5R,: (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 B relative 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 B fall from rest toward us. From this reference frame, again use Kf Uf-Ki U, to find the following when the center-to-center separation is 0.5R: (e) the kinetic energy of body B and (f) the speed of body B relative to body A. (g) Are the answers to (d) and (f) different? If so, which answer is correct?

Explanation / Answer

a )

m = 2.4 X 1027 kg

R = 6.5 X 109 m

KEtotal = gm2 / R2

= 6.67 X 10-11 X 2.4 X 2.4 X 1027x2 / 6.5 X 109

KEtotal = 5.91 X 1034 J

b )

KE1 = KE2 = KEtotal / 2

because the mass is same

KE1 = KE2 = KEtotal / 2

KE1 = KE2 = 5.91 X 1034 / 2

KE1 = KE2 = 2.955 X 1034 J

c )

KE1 = 0.5 m v2

2.955 X 1034 = 0.5 X 2.4 X 1027 X v2

v2 = 24625000

v = 4962.35 m/s

d)

Vrelative = 2 v

= 2 X 4962.35

Vrelative = 9924.71 m/s

e )

from non inertial frame

is also same

KEB = 5.91 X 1034 J  

f )

KEB = 0.5 m VB2

5.91 X 1034 = 0.5 X 2.4 X 1027 X VB2

VB2 = 49250000

VB = 7017.83 m/s

the answers both are different.

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