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
Is it length and breadth?
But if it is then how do they make circular motion ?
I’m confused
Can someone help
Oh wait is it force and motion?
PLS SOMEONE HELP ME I've been reading and I still can’t find anything
You don’t even have to give me the answer
Can you just tell me if I’m right or wrong pls
Or is it inertia and velocity ?
Dude
I have no clue
Oh wait the question at the top is not the one I meant to ask
This is:
Describe two dimensions of the motion of an object in a circle due to centripetal force. Explain why putting them together results in circular motion
I don’t care about the question at the very top. I’ve already got that one. It’s the other question I care about
I just need to know the two dimensions that cause an object to move in a circle and why
To determine the area swept by the radius vector during a certain time period, we can use the concept of the conservation of angular momentum. According to this principle, the planet's orbital angular momentum remains constant throughout its orbit.
Given that the radius vector sweeps an area A in 45 days, we can calculate the orbital angular momentum during this period.
The orbital angular momentum (L) is given by the equation:
L = mvr,
where m is the mass of the planet, v is its orbital speed, and r is the radial distance.
Now, let's consider the scenario when the planet is at its slowest orbital speed. Since the orbital angular momentum is conserved, it implies that when the speed is slowest, the radius vector must be longest.
Let's define the initial radius vector as r1 and the corresponding orbital speed as v1. Thus, the initial angular momentum is L1 = m * v1 * r1.
Next, let's consider the scenario when the planet is at its fastest orbital speed. Since the orbital angular momentum remains the same, the speed will be at its maximum when the radius vector is shortest.
Let's define the final radius vector as r2 and the corresponding orbital speed as v2. Thus, the final angular momentum is L2 = m * v2 * r2.
Since the angular momentum is conserved, we have L1 = L2.
Expanding the equation, we get:
m * v1 * r1 = m * v2 * r2.
Dividing both sides by m and rearranging the equation, we have:
v1 * r1 = v2 * r2.
Now, suppose the radius vector sweeps an area X during a 40-day time period while at its fastest orbital speed. We want to compare X with A.
The area swept, X, is given by the equation:
X = 0.5 * r2 * v2 * t,
where t is the time period, in this case, 40 days.
Substituting 0.5 * r2 * v2 from the equation above, we get:
X = 0.5 * (v1 * r1) * t.
Since we know that v1 * r1 = v2 * r2, the equation becomes:
X = 0.5 * (v2 * r2) * t = 0.5 * L2 = 0.5 * (m * v2 * r2).
Therefore, the area X swept during the 40-day time period while at the fastest orbital speed is 0.5 times the area A swept during the 45-day time period while at the slowest orbital speed.
Thus, the answer is C) 0.5 A.