Question

Consider the Sun to be at the origin of an xy-coordinate system. A telescope spots an asteroid in the xy-plane at a position given by (1.8 1011 m, 3.1 1011 m) with a velocity given by (-8.4 103 m/s, -6.3 103 m/s). What will be the magnitude of the asteroid's velocity and it's distance from the Sun at closest approach?

Answer 1

This is a very clever problem. Here is how to do it.

The cross product of the V and R vectors is the rate at which area is swept out by the asteroid, which (according to Kepler's second law) is a constant related to the angular momentum. Compute its value using the experimental measurements.

C1 = |R|*Vtheta
= [(3.1)(8.4) - (6.3)(1.8)]*10^11
= 14.7*10^11 m^2/s

Vtheta is the velocity component perpendicular to the position vector, R

Another constant of motion is the total energy, which is (1/2)mV^2 -GMm/R.
Since m (the asteroid mass) is constant,
V^2/2 - GM/R = C2 , another constant. Compute this constant using G, M(solar) and the observed values of the magnitudes of R and V.

At perihelion, the C1 constant is the product of the velocity Vmax and closest approach distance, Rmin

Vmax*Rmin = C1
Vmax^2/2 - G*M/Rmin = C2

You now have two equations for the two unknowns, Vmax and Rmin. You will need to use the other known constants, M and G

Answer 2

Could u rewrite the eqn that you used for C2

Answer 3

(V^2/2) - (GM/R) = C2

(Vmax^2/2) - (G*M/Rmin) = C2

It is the sum of kinetic and potential energy per unit mass. Using the first equation and the |V| and |R| magnitudes of the observation, you can compute the value of C2.

Answer 4

i am unable to get the right answer..please help

Answer 5

Sorry. You'll have to fill in the blanks. I have explained how to do it. Make sure you are using the correct values of G and M, and that your equations are dimensionally consistent.

Answer 6

what drwls did makes no sense..