# when a planet is at its slowest orbital speed its radius vector sweeps an area, A, in 45 days. what area will the radius vector for this planet sweep during a 40-day time period while at its fastest orbital speed?

A) A
B) More than A
C) 0.5 A
D) Less than A

I need help pls someone 😭 I can’t help that I’m stupid

## Less than A

Just took the test, this is the correct answer

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## Oh, dear friend, you're not stupid! Don't worry, I'll lend you a humorous hand to crack this orbital puzzle for you!

Now, let's think about it: when a planet is at its slowest orbital speed, it takes 45 days to sweep an area A. Great, we have that as our starting point!

Now, imagine the planet is revving up its engines and moving at its fastest orbital speed for a 40-day period. Vroom, vroom!

Since it's moving faster, it will cover more ground in 40 days. Well, guess what? The area swept by the radius vector during this time period will be...wait for it...B) More than A!

Yes, you heard it right. When the planet is at its fastest orbital speed, the radius vector will actually sweep an area greater than A. That's planetary efficiency for you! So, choose option B and pat yourself on the back. You've got this!

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## You are not stupid! I'm here to help you. Let's break down the problem step by step.

We have a planet that has a slowest orbital speed. This means that the planet's fastest orbital speed will occur at a different time. During the slowest orbital speed, the radius vector sweeps an area A in 45 days.

To determine the area that the radius vector will sweep during a 40-day time period when the planet is at its fastest orbital speed, we can use the concept of the conservation of angular momentum.

The formula for the conservation of angular momentum is:

r1 x m1 x v1 = r2 x m2 x v2

Where:
m1 = initial mass (which remains constant)
v1 = initial velocity (slowest orbital speed)
m2 = final mass (which remains constant)
v2 = final velocity (fastest orbital speed)

Since the mass remains constant, we can cancel out m1 and m2:

r1 x v1 = r2 x v2

Now, we are given that the radius vector sweeps an area A in 45 days at the slowest orbital speed. We can rewrite the equation as:

r1 x v1 x 45 = A

To find the area (A2) that the radius vector will sweep during a 40-day time period at the fastest orbital speed, we rearrange the equation:

r2 x v2 x 40 = A2

Now we can substitute the values:

r1 x v1 x 45 = A
r2 x v2 x 40 = A2

Divide the two equations to get:

(r1 x v1 x 45) / (r2 x v2 x 40) = A / A2

Since the mass remains constant, we know that r1 = r2, so we can simplify the equation to:

(45 x v1) / (40 x v2) = A / A2

Since the orbital speed is inversely proportional to the time it takes to complete one orbit, we know that v1 < v2, and therefore, (45 x v1) < (40 x v2).

Thus, A / A2 < 1. This means that the area (A2) that the radius vector will sweep during a 40-day time period at the fastest orbital speed will be less than the area (A) swept during the 45-day time period at the slowest orbital speed.

D) Less than A

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## You're not stupid! I'm here to help you understand this problem.

To determine the area that the radius vector of a planet will sweep during a certain time period, we can use the concept of the conservation of angular momentum. According to this concept, when an object is moving in a circular path, its angular momentum remains constant unless acted upon by external forces.

In this case, we are given that the radius vector sweeps an area A in 45 days when the planet is at its slowest orbital speed. We can denote the initial angular momentum as L1 and the initial area as A1.

Now, we need to determine the area that the radius vector will sweep during a 40-day time period when the planet is at its fastest orbital speed. Let's denote this area as A2.

Since angular momentum is conserved, we have:

L1 = L2

Now, angular momentum is given by the equation:

L = m * r * v

where m is the mass of the planet, r is the radius of the circular path traced by the radius vector, and v is the orbital speed of the planet.

Since the mass of the planet and the radius of the circular path remain constant, we can rewrite the equation as:

r * v = constant

Now, from the given information, when the planet is at its slowest orbital speed, the radius vector sweeps an area A in 45 days. So, we have:

A1 = r1 * v1 * t1

where r1 is the radius of the circular path, v1 is the slowest orbital speed, and t1 is the time period (45 days).

Similarly, when the planet is at its fastest orbital speed, we have:

A2 = r2 * v2 * t2

where r2 is the radius of the circular path, v2 is the fastest orbital speed, and t2 is the time period (40 days).

Now, we need to find the relationship between A2 and A1. Dividing A2 by A1, we have:

A2/A1 = (r2 * v2 * t2) / (r1 * v1 * t1)

Since we are comparing the fastest and slowest orbital speeds, we can assume that the radius of the circular path remains constant. Therefore, r1 = r2, and we can cancel out this term from the equation:

A2/A1 = (v2 * t2) / (v1 * t1)

Now, if the planet is at its slowest orbital speed during the sweep of area A1, it means that it takes less time to traverse that area compared to 45 days. Similarly, if the planet is at its fastest orbital speed during the sweep of area A2, it will take less time to traverse that area compared to 40 days.

Therefore, we can conclude that the ratio of the time period for the fastest speed (t2) to the time period for the slowest speed (t1) is less than 1:

t2/t1 < 1

Since the ratio of the time periods is less than 1, the ratio of the areas swept by the radius vector will also be less than 1:

A2/A1 < 1

Hence, the answer to the question is (D) Less than A.