Ameteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true anomaly of 150. if the speed of the meteoroid at that time is 2.23 km/s, calculate
(a) the eccentricity of the trajectory;
(b) the altitude at closest approach; and
(c) the speed at the closest approach.

Explanation:

Given that,

Distance = 402000 km

Speed = 2.23 m/s

Angle = 150

(a). We need to calculate the eccentricity of the trajectory

Using formula of eccentricity

Put the value into the formula

We need to calculate the angular momentum

Using formula of  the angular momentum

...(I)

The orbit equation is

....(II)

Equating the value of h²

(b). We need to calculate the altitude at closest approach

Put the value of e in equation (I)

Now, using the formula of the altitude at closest

So, The altitude is

(c). We need to calculate the  speed at the closest approach.

Using formula of speed

Hence, This is the required solution.

what is the actual question though?

the right answer is 0.3 meters per second