Angol nyelvű szám, 2002. december
	Előző oldal	Tartalomjegyzék	Következő oldal	MEGRENDELŐLAP

Solutions of problems for physics P

P. 3436. One of the planets of star Noname is long and cylindrical. The average density of the planet is identical to that of the Earth, its radius equals that of the Earth and the period of its rotation is exactly 1 day.

a) What is the first cosmic speed for this planet?

b) How high above the surface of the planet do the synchronous telecommunications satellites orbit?

c) What is the second cosmic speed for this planet?

(5 points)

Submitted by: Horányi Gábor, Budapest

Solution. The gravitational field around a very long cylinder shows cylindrical symmetry, and the direction of the field strength (sufficiently far from the ends) is radial, which means that it is perpendicular to the axis of the cylinder and its magnitude depends solely on the distance from that axis.

Using the analogy of gravitational and Coulomb fields we can make the statement that the number of the gravitational field lines (the product of the g gravitational acceleration and the surface perpendicular to it) `leaving' the body of mass m is 4\(\displaystyle \pi\)f ^.m where f is the Newtonian gravitation constant. Therefore, the connection between the g-lines leaving a cylinder of length L and radius r and the mass inside the cylinder is:

g^.2r\(\displaystyle \pi\)L=4\(\displaystyle \pi\)f R²\(\displaystyle \pi\)L\(\displaystyle \varrho\), that is \(\displaystyle g(r)=\frac{2\pi fR^2\varrho}{r}\).

a) According to the above law of force (and motion equation mg=mv²/r) the velocity of a body revolving in a circular orbit, independently of the radius of the orbit (so on the surface of the planet as well), is:

\(\displaystyle v=\sqrt{2fR^2\pi\varrho}, \)

accordingly, this is the value of the first cosmic speed. This is \(\displaystyle \sqrt{3/2}\) times higher than the value valid for the Earth

\(\displaystyle v_F=\sqrt{\frac{fM_{\rm Earth}}{R}}=\sqrt{ \frac{4}{3}R^2\pi f\varrho}=7.9~\rm km/s \)

that is about 9,7 km/s.

b) The period of a satellite revolving in an orbit of radius r is T_r=2\(\displaystyle \pi\)r/v. If one `day' is T₀ long, then the orbiting radius of a synchronous satellite is

\(\displaystyle r_0=\frac{T_0v}{2\pi}=R\sqrt{\frac{T_0^2f\varrho}{2\pi}}. \)

In the case of the Earth this radius is \(\displaystyle r^*=R{\root3\of{T_0^2f\varrho/3\pi}}\), that is

\(\displaystyle r_0=\sqrt{\frac{2{r^*}^3}{3R}}\approx1.33\cdot10^8~\rm m. \)

Synchronous telecommunications satellites therefore orbit at an

r₀-R=1.27^.10⁸ m

height above the surface of the cylinder shaped planet.

c) The second cosmic speed, which is the escape velocity from the planet, is very high, and its exact value depends on the length of the planet. If it is endless, the escape velocity is also endless because with finite energy there is no escape from a force field, which decreases in the order of the 1/r law. To prove this, let us consider a series of distances in geometrical progression: r_n=\(\displaystyle \alpha\)ⁿr₀ (where \(\displaystyle \alpha\)>1 and r₀=R). The energy \(\displaystyle E(r_{n-1}\to r_n)\) that is needed to get from the height of r_n-1 to the height of r_n is independent of n, that is with increasing n the force decreases the same rate as the distance increases. The required energy to get from r₀ to r_N is \(\displaystyle E(r_0\to r_N)=NE(r_0\to r_1)\). It is clear even from this that while putting in finite energy one can get only to a finite height.

If the planet is not endless (its length is H), then as long as we are far from its ends and r\(\displaystyle \ll\)H, the 1/r force law holds, but if \(\displaystyle r\simeq H\), the characteristics of the force law change, and if r\(\displaystyle \gg\)H the usual 1/r² rule takes over. From a planet like this an escape is possible at a certain finite velocity.

Based on the papers of
Dolgos Gergely (11th form student of Árpád Secondary Grammar School, Budapest) and
Siroki László (11th form student of Fazekas M. Secondary Grammar School, Debrecen)

Note. Using the means of the integral calculus it can be shown that the second cosmic speed equals the \(\displaystyle \sqrt{2}\cdot\ln\,(H/R)\)-fold value of the first cosmic speed (In Earth's case it is \(\displaystyle \sqrt{2}\).) This logarithmic factor doesn't get too big even if H\(\displaystyle \gg\)R (when H=10 R v_II/v_I is about 3, and at the value of H=1000 R it is still lower than 10).

(G. P.)