USAPhO 1998, Четвъртфинал — Задача 4

• (a, 7) $C=4\pi^2/GM$
• (b, 6) $4.22 \times 10^{7}$ m
• (c, 6) -6.1 s
• (d, 6) 3.3 cm/s east.

(a, 7) $\mathbf F=m\mathbf a$ applied to the satellite gives

$GMm/r^2=mv^2/r$ (3)

where for the satellite, $v=2\pi r/T$. (3)

Thus one obtains $T^2=(4\pi^2/GM)r^3$,

so that $C=4\pi^2/GM$. (1)

(b, 6) For geosynchronous orbit, $T=24\ hrs.=8.64\times10^4\ s$ (3)

Thus Kepler’s third law and the Earth data give $r=4.22\times10^7\ m$. (3)

(c, 6) Let $T_0=8.64\times10^4\ s$, and let $r_0=4.22\times10^7\ m$, i.e., let the subscript denote the geosynchronous orbit data. With the change in orbit radius, we have $r=r_0+\Delta r$ and $T=T_0+\Delta T$, so Kepler’s third law becomes

(1)

Use the binomial expansion to obtain

$\Delta T=(3/2)(T_0/r_0)\Delta r$. (3)

With $\Delta r=-2\ km$, it follows that $\Delta T=-6.1\ s$. (2)

[Of these last two points, one point is for the 6.1 s; one point is for either including the minus sign or stating that the satellite’s period is shorter than before.]

(d, 6) Both the satellite and observer are revolving about the Earth’s center of mass.

Let $v_s$ = speed of the satellite relative to Earth’s center of mass;

Let $v_o$ = speed of the observer relative to Earth’s center of mass.

The quantity we seek is $v_s-v_o$. Thus,

(2)

Writing $T_s=T_o(1+\Delta T/T_o)$ and using the binomial expansion, one finds

$v_s-v_o=-(2\pi R_e/T_o^2)\Delta T$. (2)

Numerically, $v_s-v_o=-2\pi(6.37\times10^6\ m)(8.64\times10^4\ s)^{-2}(-6.1\ s)$

$=3.3\ cm/s$. (1)

Because the Earth rotates from west to east, and because the satellite completes one “lap” about the Earth’s C.M. in 6.1 s less than the observer, the spot moves east of the observer at the speed of 3.3 cm/s. (1)