the more straightforward:

Assume perfectly circular orbits, and that the two planets start at

conjunction.

let A_e be the angular position of the earth,

A be the angular position of the planet.

T_e is the sidereal period of the earth,

T is the sidereal period of the planet

At any time, t, the angular position of the earth as measured from

conjunction is:

A_e(t) = 2*pi*t/T_e. Likewise for the planet:

A(t) = 2*pi*t/T.

i) An inferior planet will completely circle the sun and reach the

next conjunction with earth before the earth has completed one sidereal

period. The planet will have travelled exactly 2*pi further than the earth

has, so that at the next conjunction, t=S, and

A_e(S) = A(S) - 2*pi, so that:

2*pi*S/T_e = 2*pi*S/T - 2*pi. Solving for S gives:

S(1/T - 1/T_e) = 1, so that 1/S = 1/T - 1/T_e.

ii) The earth will completely circle the sun and reach conjunction with a

superior planet before that planet has completed a period. So, analogous to

the above:

A(S) = A_e(S) - 2*pi, so that solving for S gives:

S(1/T_e - 1/T) = 1, and 1/S = 1/T_e - 1/T

S_v = 583.9 day

S_m = 779.9 day

P_e = (sidereal period of earth) = 365.5 day

venus: P_v = (S_v*P_e)/(S_v+P_e) = 224 day

mars : P_m = (S_m*P_e)/(S_m+P_e) = 687 day

b) Mercury has the shortest synodic period because its small sidereal period

enables it to lap the earth quicky. Mercury's sidereal period is 0.2408yr,

so that sometime between 0.2408yr and 0.4816yr, it will again

be in conjunction with the earth. Again using equation 1.1 for Mercury

shows that its synodic period is 0.312 yr.

The second most popular answer was Pluto, but Pluto can't have a smaller

synodic period than Mercury because the earth will take a little more than

one year to lap Pluto. Pluto does have the smallest synodic period

amongst the superior planets, but Mercury has the smallest among all

the planets.

For the sun: | RA(hour) | dec (degrees) |

Vernal equinox: | 0 | 0 |

Summer solstice: | 6 | 23.5 |

Autumnal equinox: | 12 | 0 |

Winter solstice: | 18 | -23.5 |

The declination of 23.5 is the angle between the earth's equator and

the ecliptic.

is:

h = 90 - (L - dec)

i) summer solstice: The sun's declination, dec, is 23.5 degrees,

so, at latitude 42 degrees,

h = 90 - (42 - 23.5) = 71.5 degrees

ii) winter solstice: The sun's declination, = -23.5 degrees,

so,

h = 90 - (42 + 23.5) = 24.5 degrees

a) As shown in the above figure, at latitude L, the north pole or declination of

90 degrees lies at an altitude of L degrees above the horizon. The

zenith lies where the declination is L, and horizon lies at

declination (90-L) degrees. Always visible circumpolar stars will

never set, and these are the stars whose declinations lie between

90 degrees and (90-L) degrees (if L is greater than 0). If you're in the southern hemisphere,

with L<0, The declination of the (south) celestial pole is -90 degrees, and stars with declinations

between -90 degrees and (-90-L) degrees will always be visible.

b) At summer solstice, the sun lies at declination 23.5 degrees. Using the

result from part a, the sun never sets in those latitudes

where the declination > 90 -L, or,

23.5 degrees > 90 -L, so that

L > 90 - 23.5 = 66.5 degrees, i.e. for latitudes within the arctic circle.

c) at vernal equinox, the sun lies essentially at declination 0 degrees.

Now, the equinox means that every point on earth will experience equal

amounts of darkness and sunlight that day. So, that implies that there

is no place on earth where the sun never sets. Strictly speaking, it

can be argued that the poles ARE the only place where the sun never sets,

because there, the sun just orbits the horizon. So, the sun never rises

and consequently never sets.

You can also use part a) to argue that no latitude can be found

which satisfies 0 > 90 -L.