# Asteroid close encounter: speed at perigee

1. May 13, 2017

### rickyjoepr

1. The problem statement, all variables and given/known data
an asteroid heading towards earth, has a speed of 7.25 km/s as it passes the moons orbit, kit misses earth by 5000km. What is the speed at the closest point to earth?

The professor provided us with the solution, which should be 11.0 km/s, however when I do the calculation, using the given values, my number is way off.

Is there a conversion I should be doing that I may not know about?

2. Relevant equations
Conservation of energy for an asteroid is
(.5) mVi^2-((GmMearth)/Ri) = (.5) mVf^2-((GmM_earth)/Rf)

3. The attempt at a solution
Vi = 7.25 km/s

Ri = distance from the moon to earth = 384,000km
Rf = radius of earth + 5000km = 11400km
Mearth =5.972x10^24

G = 6.67384x10^-11

solve for Vf

Vf = SQRT[ Vi^2 + 2GMearth ((1/Rf)-(1/Ri))] = 260475.0801

2. May 13, 2017

In your formulas, you use kilograms which is correct, but you need to use meters and not kilometers for distance as well as velocity (m/sec). The constant $G$ works in meters, and the energy it computes works in m/sec. Once you solve for $v_f$ in m/sec, you can then convert it back to km/sec.

3. May 13, 2017

### rickyjoepr

I thought it was an issue with G, I was trying to convert G to km , but I will try what you suggested

edit: and even if i dd convert G to km, it would still be wrong as the kg Mass of earth must work with m/s

4. May 13, 2017

The energy that your equation $U=-GMm/r$ computes is in joules, so it is much easier to work with $G$ in the M.K.S. system as is, and simply convert the velocities and distances, and convert $v_f$ back at the end to km/sec.