0.1 Problem 1

A satellite is in an orbit with a period $T = 205$ minutes and eccentricity $e = 0.40$ about the Earth. When the true anomaly of the satellite is $f = 70$ degrees, ﬁnd the time $t − τ$ since perigee passage, in minutes.

Answer

n (t − τ) = E − esin E

But $n = 2Tπ$ hence

t− τ = E-−--esin-E-= T-(E-−--esin-E)- n 2π

(1)

But $∘ ---- ( f) 1+e (E-) tan 2 = 1−e tan 2$ , hence $E$ can be found. Substituting it in the above, solves for $t − τ$

Hence from Eq (1)

0.2 Problem 2

A satellite is in an orbit with a period $T = 205$ minutes and eccentricity $e = 0.40$ about the Earth. Find the true anomaly of the satellite, in degrees, when it is $50$ minutes past perigee passage.

Answer

Solving for $E$

E = 1.9097 rad

Hence

0.3 Problem 3

A spaceship in a circular orbit above the Earth at an altitude of $300$ km. At time $t = 0$ , it retroﬁres its engine, reducing its speed by $500$ m/s. How long (in minutes) does it take to impact the Earth? Neglect atmospheric drag.

Answer

pict

μ = 3.986× 105 km3/s2

But

ΔV = V2 − V1

Where $∘ --- ∘ ------ V1 = rμ = r-μ+alt a E$ where $rE$ is earth radius and $alt$ is the altitude at $t = 0$ when the spaceship was in circular orbit. Hence $∘ -------5 V1 = 36.938768×+13000-= 7.7258$ km/s hence $V2 = V1 − 500 × 10−3 = 7.7258 − 0.5 =$ $7.2258$ km/sec. This is the speed at apogee for the new orbit.

Va = 7.2258 km/sec

But

But also we know that $ra = a (1+ e)$ , hence

Substitute (2) in (1)

Hence from (2) we ﬁnd $a$

---6678---- a = 1+ 0.12526 = 5934.6

Hence $n$ the mean speed is

At impact $r = rE$ , hence

Solving this equation gives $E = 1260$ , but we want to use the $E$ shown in the diagram. Hence remember to do $E = 2π − E actual$ to obtain $E = 233o actual$ and that is the $E$ to use in $n(t− τ) = E − e sin E.$ Hence, measured from perigee,