## Task 1 - 030911

The first female cosmonaut in the world, Valentina Tereshkova, launched with her spacecraft Vostok 6 on June 16, 1963, at 10:30 AM and landed after 48 orbits of the Earth on June 19, 1963, at 9:20 AM. The average flight altitude was $200 \mathrm{~km}$. (Average Earth radius $R=6370 \mathrm{~km}$.)

a) How many kilometers did the cosmonaut travel during her space flight? (For simplification, assume that the launch and landing sites coincided and that the flight took place on a circular orbit.)

b) What was the average speed during the space flight?
a) Given are the mean Earth radius $R=6370 \mathrm{~km}$, the average flight altitude $h=200 \mathrm{~km}$, and the number $n$ of Earth orbits, $n=48$. The orbit is assumed to be circular, with the takeoff and landing points being identical.

The total flight distance $s$ consists of three segments: the takeoff flight distance $s_{S}$ to reach the flight altitude after takeoff, the flight distance $s_{U}$ covered during the Earth orbits, and the landing flight distance $s_{L}$ to leave the flight altitude before landing.

We will assume for simplicity that the flight altitude is reached and left vertically above the takeoff and landing points, respectively. It then follows that $s_{S}=s_{L}=h$.

To calculate the distance $s_{U}$, it is necessary to know the circumference $u$ of the orbit - which is obviously $s_{U}=n u$. The circumference of a circle can be determined from its radius $r$; the radius of the orbit is, of course, the sum of the Earth radius and the flight altitude, $r=R+h$.

Thus, the total flight distance $s$ is calculated as follows:

$$
\begin{aligned}
s & =s_{S}+s_{U}+s_{L}=h+n u+h \\
& =2 h+2 n \pi r=2 h+2 n \pi(R+h) \\
& =2 \cdot 200 \mathrm{~km}+2 \cdot 48 \pi(6370 \mathrm{~km}+200 \mathrm{~km})=1982000 \mathrm{~km}
\end{aligned}
$$

b) The flight began on June 16 at 10:30 AM and ended on June 19 at 9:20 AM; it thus lasted 2 days, 22 hours, and 50 minutes. The flight time $t$ is therefore 4250 minutes.

With $s=1982000 \mathrm{~km}$ according to (a), the average flight speed $v$ is:

$$
v=\frac{s}{t}=\frac{1982000 \mathrm{~km}}{4250 \mathrm{~min}}=466 \frac{\mathrm{km}}{\min }=27980 \frac{\mathrm{km}}{\mathrm{h}}=7.77 \frac{\mathrm{km}}{\mathrm{s}}
$$