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
Explanation:
<h2>Part a </h2>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.
<h2>Part b</h2>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
<h2>Part c</h2>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
<h2>Part d</h2>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