Question

The asteroid Icarus has an elliptical orbit that carries it between 0.19 and 1.97 AU from the Sun. What is its semimajor axis? How often does it cross Earth’s orbital radius? Its semimajor axis is AU. (Round the final answer to two decimal places.) It crosses the Earths orbital radius every how many days on average? (Round the final answer to the nearest whole number.)

Answer 1

To find the semimajor axis of the asteroid Icarus, we can use the formula:

Semimajor axis (a) = (Perihelion distance + Aphelion distance) / 2

Given that the perihelion distance is 0.19 AU and the aphelion distance is 1.97 AU, we can substitute these values into the formula:

Semimajor axis (a) = (0.19 AU + 1.97 AU) / 2 = 2.16 AU

So the semimajor axis of the asteroid Icarus is approximately 2.16 AU.

To determine how often it crosses Earth's orbital radius, we need to calculate the period of Icarus' orbit. The period can be approximated using Kepler's third law:

Period (T) = 2π * √(a^3 / G * M)

Where:
- T is the period of the orbit.
- a is the semimajor axis.
- G is the gravitational constant (approximately 6.67430 × 10^(-11) m^3 kg^(-1) s^(-2)).
- M is the solar mass (approximately 1.989 × 10^30 kg).

We'll convert the semimajor axis from AU to meters by using the conversion factor: 1 AU = 1.496 × 10^11 meters.

a = 2.16 AU * 1.496 × 10^11 m/AU = 3.23056 × 10^11 meters

Substituting the values into the formula:

Period (T) = 2π * √((3.23056 × 10^11)^3 / (6.67430 × 10^(-11) * 1.989 × 10^30))

Calculating this expression gives us T ≈ 822.81 days.

Therefore, the asteroid Icarus crosses Earth's orbital radius approximately every 822 days on average.

Answer 2

The semimajor axis of the asteroid Icarus' elliptical orbit is 1.08 AU. It crosses Earth's orbital radius twice during each orbit.

Step-by-step explanation:

The semimajor axis of an elliptical orbit is the average distance between the orbit and the focus (in this case, the Sun). To find the semimajor axis, we can use the formula:

Semimajor axis = (Perihelion distance + Aphelion distance) / 2

For the asteroid Icarus, the perihelion distance is 0.19 AU and the aphelion distance is 1.97 AU. Plugging these values into the formula, we get:

Semimajor axis = (0.19 + 1.97) / 2 = 1.08 AU

Therefore, the semimajor axis of Icarus' orbit is 1.08 AU.

To calculate how often Icarus crosses Earth's orbital radius, we need to compare the semimajor axis of Icarus with Earth's semimajor axis, which is approximately 1 AU. Since Icarus' semimajor axis is 1.08 AU, it crosses Earth's orbital radius twice during each orbit. To find the average time between crossings, we need to know the orbital period of Icarus. Unfortunately, the question does not provide this information, so we cannot calculate the exact number of days between crossings.