Answer:
Part a: The eccentricity is 1.086.
Part b: The altitude at closest approach is 5088 km
Part c: The velocity at perigee is 8.516 km/s
Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km
Step-by-step explanation:
Part a
Specific energy is given by

Here
- ε is the specific energy
- v is the velocity which is given as 2.23 km/s
- μ is the gravitational constant whose value is 398600
- r is the distance between earth and the meteorite which is 402,000 km

Value of specific energy is also given as

Orbit formula is given as

Putting values in this equation and solving for e via the quadratic formula gives

As the value of eccentricity cannot be negative so the eccentricity is 1.086.
Part b
The radius of trajectory at perigee is given as

Substituting values gives

Now for estimation of altitude z above earth is given as

So the altitude at closest approach is 5088 km
Part c
radius of perigee is also given as

Rearranging this equation gives

Now the velocity at perigee is given as

So the velocity at perigee is 8.516 km/s
Part d
Turn angle is given as

Substituting value in the equation gives

Aiming radius is given as

Substituting value in the equation gives

So the turn angle is 134.08 while the aiming radius is 5641.28 km