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?

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

Could u rewrite the eqn that you used for C2

(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.

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

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.

what drwls did makes no sense..

To find the magnitude of the asteroid's velocity, we can use the Pythagorean theorem. The magnitude of the velocity vector is given by the square root of the sum of the squares of its components. In this case, the components of the velocity are (-8.4 x 10^3 m/s, -6.3 x 10^3 m/s).

The magnitude of velocity (v) can be calculated as:

v = √(vx^2 + vy^2)

where vx is the x-component of velocity and vy is the y-component of velocity.

Substituting the given values, we have:

v = √((-8.4 x 10^3 m/s)^2 + (-6.3 x 10^3 m/s)^2)

Now, we just need to evaluate this expression:

v = √(70.56 x 10^6 m^2/s^2 + 39.69 x 10^6 m^2/s^2)

v = √(110.25 x 10^6 m^2/s^2)

v = √(110.25) x 10^3 m/s

v = 10.5 x 10^3 m/s or 1.05 x 10^4 m/s

So, the magnitude of the asteroid's velocity is 1.05 x 10^4 m/s.

To find the distance from the Sun at closest approach, we need to find the distance between the Sun (at the origin) and the position of the asteroid (1.8 x 10^11 m, 3.1 x 10^11 m).

The distance (d) between two points in a two-dimensional Cartesian coordinate system can be calculated using the distance formula:

d = √((x2 - x1)^2 + (y2 - y1)^2)

where (x1, y1) are the coordinates of the first point (in this case, the origin) and (x2, y2) are the coordinates of the second point (the position of the asteroid).

Substituting the given values, we have:

d = √((1.8 x 10^11 m - 0)^2 + (3.1 x 10^11 m - 0)^2)

Now, we just need to evaluate this expression:

d = √((1.8 x 10^11 m)^2 + (3.1 x 10^11 m)^2)

d = √(3.24 x 10^22 m^2 + 9.61 x 10^22 m^2)

d = √(12.85 x 10^22 m^2)

d = √(12.85) x 10^11 m

d = 3.59 x 10^11 m

So, the distance from the Sun at closest approach is 3.59 x 10^11 m.