Question

You are an astronomer on planet tirth, which orbits a distant star. it has recently been accepted that tirth is spherical in shape, though no one knows its size. one day, you learn that on the equinox your sun is directly overhead in the city of tyene, located 750 kilometers due north of you. on the equinox, you go outside in alectown and observe that the altitude of your sun is 87.0 ∘.

Answer 1

The altitude of the Sun at noon on December 21, as seen from a place on the Tropic of Cancer, is 44°.

Step-by-step explanation:

To determine the altitude of the Sun on December 21 as seen from a place on the Tropic of Cancer, we need to consider the tilt of Earth's axis. On this date, the Sun is located about 23° south of the celestial equator. From the Tropic of Cancer, which is at a latitude of 23° north, the zenith would be at a declination of 23° north. The difference in declination between the zenith and the position of the Sun is 46°, so the Sun would be 46° away from the zenith. Therefore, its altitude would be 90° - 46° = 44°.

Answer 2

The circumference of planet tirth is
`C = 90000 \ km`

Step-by-step explanation:

From the question we are told that

The distance of tyene from the observer is
`d = 750 \ km`

The altitude of the sun from the equinox is
`\theta_s = 87.0^o`

Generally given that on the equinox your sun is directly overhead in the city of tyene , then the altitude of the tyene from the equinox is
`\theta_e = 90 ^o`

Generally the angular distance between tyene and the sun as observed from the equinox is mathematically represented as

`\theta _d = \theta_e - \theta_s`

=>
`\theta _d = 90 -87`

=>
`\theta _d =3^o`

Generally the ratio of
`\theta _d =3^o` to
`360^o` which is the total angle in a circle is equal to the ratio of
`d = 750 \ km` to the circumference of the planet

So

`(\theta_d)/(360) = (750)/(C)`

=>
`(3)/(360) = (750)/(C)`

=>
`C = 90000 \ km`