On March 18 astronomers noticed unknown asteroid right next to Polaris?

Question

On March 19 the asteroid deviated from Polaris only slightly, while its visible size increased considerably. The astronomers announced that doomsday arrived.

Fortuantely astronomers were wrong. On March 20 the asteroid missed the Earth by mere 100 km.

Two days later astronomers said that right now they could not track the asteroid, because it was passing the bright disk of Sun, and was invisible.

What was velocity of the astroid with respect to the center of Earth at the closest approach?

This problem frakly is heavy on trivial knowledge, nemely:

At equinox the angle between axis or Earth and Sun is π/2.
Asteroids follow hyperbolic orbits with Earth at one focus.

Armed with this knowledge we conclude that asteroid was scattered at angle π/2, and therefore followed tragectory most easily described as xy=1. The focus of this hyperbola can be also easily placed at (√2, √2).

Angular momentum at infity was
L = Vo √2, and energy
E = 1/2 Vo².
Angular momentum at closest approach (1,1) [which by the way happened at 45deg S] was
L = Vmax (2-√2), and energy
E = 1/2 (Vmax² - Vesc²).

Equations:
Vmax(√2-1) = Vo
Vmax² - Vesc² = Vo²

Vmax² - Vesc² = Vmax²(√2-1)²
Vmax² - Vesc² = Vmax²(3 - 2√2)
Vmax² = Vesc²/(2√2 - 2)
Vmax = Vesc /√(2√2 - 2) ~ 1.10 Vesc

Answer: 12.3km/s

Answer 1

At least with respect to the earth, our little asteroid has a hyperbolic orbit.

On March 20, the north pole will be pointing in a direction perpendicular to the sun therefore the angle change of the asteroid ~ 90 deg

The eccentricity (e) for a theta of 45 degrees (180 - 2*theta= 90degrees) is going to be 1/sin (theta) = squrt 2

a=(e-1)rp

Perihelian distance rp= Earth radius + 100k
=6370 +100
=6470 km

v = sqr(mu*(2/r+1/a)) ref: http://en.wikipedia.org/wiki/Orbital_velocity

mu= standard gravitational parameter = GMearth= g*r^2

(I know, I'm about to cheat and use g at 9.81 m/sec instead of computing it at a 'mere' 100 k above the earth's surface).

now sub in rp for r everywhere

v = sqr(mu*(2/r+1/a))
= sqr(g*rp^2*(2/rp+1/((e-1)rp)))
=sqr(g*rp*(2+1/(e-1)))
=sqr(9.81*6,470,000m*(2+1/(1.414-1)))
=16,740 m/sec
or
=16.74 km/sec

Answer 2

How many kilometers is it to the center of the earth?

You probably need to know:

Distance from sea level to center.
Distance from 100 km altitude to center of earth.
Whatever equation you plug everything into.

Probably a few other things.

Answer 3

Really? When did that happen again? March of what year?
Which sun? Which Earth?