PLS HELP- I NEED HELP WITH MY PHYSICS HW

To understand why, let's use Kepler's third law of planetary motion. Kepler's third law states that the square of the period of an orbiting planet is proportional to the cube of its average distance from the star.

In this case, planet 1 has an orbital radius r1, and planet 2 has r2 = 4r1. Since planet 2 has a larger orbital radius, it will take longer to complete one orbit.

According to Kepler's third law, we can write the following equation:

T1^2 = k * r1^3 (Equation 1)

T2^2 = k * r2^3 (Equation 2)

Here, k is a constant that depends on the mass of the star. Since both planets are orbiting the same star, the value of k is the same for both equations.

We can rewrite Equation 2 in terms of r1 by substituting r2 = 4r1:

T2^2 = k * (4r1)^3

Simplifying this equation:

T2^2 = 64k * r1^3

Since T1^2 = k * r1^3, we can substitute this into the equation above:

T2^2 = 64 * T1^2

Taking the square root of both sides:

T2 = 8 * T1

Therefore, the period of planet 2 (T2) is 8 times the period of planet 1 (T1).

msm555 msm555 · Answer 2 · 2024-01-05T10:30:16+01:00

Answer:

[tex]\sf \, D.\,T_2 = 8T_1[/tex]

Explanation:

The orbital period (T) of a planet is related to its orbital radius (r) by Kepler's Third Law of Planetary Motion, which states:

[tex]\boxed{\sf T^2 \propto r^3}[/tex]

This law implies that the square of the orbital period is directly proportional to the cube of the orbital radius.

For Planet 1:

[tex]\sf T_1^2 \propto r_1^3[/tex]

For Planet 2:

[tex]\sf T_2^2 \propto r_2^3[/tex]

Given that r₂ = 4r₁, we can substitute this into the equation for Planet 2:

[tex]\sf T_2^2 \propto (4r_1)^3[/tex]

Now, let's compare the two equations:

[tex]\sf T_1^2 \propto r_1^3[/tex]

[tex]\sf T_2^2 \propto 64r_1^3[/tex]

We can see that T₂ is related to T₁ by a factor of [tex]\sf \sqrt{64} = 8[/tex], because (T₂)² is proportional to [tex]\sf 64r_1^3[/tex], and the square root of 64 is 8.

So, the relationship between the orbital periods of the two planets is:

[tex]\sf T_2 = 8T_1[/tex]

Planet 2 has an orbital period that is 8 times longer than that of Planet 1.

So, the answer is:

[tex]\sf \, D.\,T_2 = 8T_1[/tex]