The obliquity of Uranus is \(98^{\circ} .\) From one of the planet's poles, how far from the zenith would the Sun appear on summer solstice?

Obliquity refers to the angle between a planet's rotational axis and its orbital plane. For Uranus, this angle is extremely large at 98 degrees. This means that Uranus rotates almost on its side compared to the plane of its orbit around the Sun. The obliquity of a planet significantly affects how it experiences seasons. For instance, while Earth's obliquity of approximately 23.5 degrees gives us moderate seasonal changes, Uranus’s severe tilt results in extreme variations where one pole can face the Sun continuously for about 42 Earth years and then be in complete darkness for another 42 years.

The summer solstice is the point in a planet’s orbit when one of its poles is tilted closest to the Sun. During this time, the pole experiences the maximum amount of daylight. For Uranus, which has an obliquity of 98 degrees, the summer solstice results in one of its poles pointing almost directly at the Sun. Imagine standing on this pole; the Sun would appear very high in the sky during the solstice. This extreme axial tilt means the Sun would be at an angle from the equatorial plane equal to the planet's obliquity at this point.

The zenith is the point in the sky directly above an observer. It's like looking straight up from where you are standing. When thinking about the position of the Sun relative to the zenith during the summer solstice on Uranus, you need to consider the axial tilt. With an obliquity of 98 degrees, the Sun will not be directly overhead. Instead, it will be at a certain angular distance from the zenith. Calculating this involves subtracting the obliquity from 180 degrees (since a straight line is 180 degrees). Therefore, on Uranus during its summer solstice, the Sun appears 82 degrees away from the zenith, making it closer to the horizon for an observer at the pole experiencing summer.