A small asteroid is moving in a circular orbit of radius R 0 about the sun. This

Question

A small asteroid is moving in a circular orbit of radius R0 about the sun. This asteroid is suddenly struck by another asteroid. (We won't worry about what happens to the second asteroid, and we'll assume that the first asteroid does not acquire a high enough velocity to escape from the sun's gravity). Immediately after the collision, the speed of the original asteroid is V0, and it is moving at an angle ? relative to the radial direction, as shown in the figure. (Figure 1) Assume that the mass of the asteroid is m and that the mass of the sun is M, and use G for the universal gravitation constant.

Since the asteroid does not reach escape velocity, it must remain in a bound orbit around the sun, which will be an ellipse. Take the following steps to find Ra and Rp, the aphelion and perihelion distance of the asteroid after the collision. (The aphelion is the point in the orbit farthest from the sun, and the perihelion is the point in the orbit closest to the sun.)

As in most orbit problems, the most fundamental principles involved are energy and angular momentum conservation.

Find a quadratic equation of the form 0=Re^2+bRe+c that relates Re to the known quantitiesL, E, m, M, and G. In the space provided, write b, the co-efficient of Re.

Explanation / Answer

given

A small asteroid is moving in a circular orbit of radius R0 about the sun.

a quadratic equation of the form 0=Re^2+bRe+c ----------------1

here in the given equation

the coefficient of the Re^2 term is one

and

remaining terms for the given quadratic equation is

as following

0 = Re2 + ( G M m / E ) X Re - L2 / 2 X ( m . E ) -------------2

comparing equation no 1 and 2

Re2 coefficient is one

b term in the equation is ( G M m / E )

c term in the equation is L2 / 2 X ( m . E )

finally Ra and Rp values are

the following

Ra = 1/2(-( G M m / E ) +( G M m / E )+ 2 (L2 / m .E))

RP = 1/2(-( G M m / E ) -( G M m / E ) + 2 L2 / m .E))

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