Question

If you want to launch a probe that orbits the Sun with a period of 4.00 years, what should be its orbital radius? Give your answer in astronomical units. The astronomical unit AU is defined as the mean distance from the Sun to the Earth 1AU=1.50Ă—1011m. Express your answer to three significant figures. Is the orbital speed of this probe greater than, less than, or equal to the orbital speed of the Earth?

Answer 1

To launch a probe into orbit around the Sun with a 4.00-year period, its orbital radius should be 2.52 AU, which means its orbital speed would be less than Earth's.

Step-by-step explanation:

If you want to launch a probe with an orbital period of 4.00 years around the Sun, you will need to calculate the orbital radius using Kepler's third law, which states that the square of the period (P) of an orbit is proportional to the cube of the semi-major axis (a) of that orbit. When P is measured in years and a is in astronomical units (AU), the relationship is given by P² = a³. With the Sun-Earth mean distance defined as 1 AU, you can determine the probe's orbital radius in AU.

To find the orbital radius a for a period P = 4.00 years, we calculate:

(4.00)² = a³
16 = a³
a = ∛(16)
a = 2.52 AU (to three significant figures)

Since the orbital speed of a body decreases with distance from the Sun, the orbital speed of this probe would be less than the orbital speed of Earth.

Answer 2

As per Kepler's law we know that

`(T_1^2)/(T_2^2) = (R_1^3)/(R_2^3)`

here we know that

Time period of earth around sun is 1 year while its orbital radius around sun is 1 AU

Now here we know that another planet has time period T = 4 years and its orbital radius is R2

now from above equation we will have

`(1^2)/(4^2) = (1^3)/(R^3)`

`(1)/(16) = (1)/(R^3)`

`R^3 = 16`

`R = 2.52 AU = 3.78 * 10^(11) m`

so above is the orbital radius