Magnitude of displacement vector

In summary: And for 9 months?Would it be sqrt((2*(1.50 X 10^11))^2 + (1.50 X 10^11)^2)?In summary, the Earth moves around the Sun in a circular orbit with a radius of 1.50 X 10^11 m at a constant speed. The position vector for 3, 6, 9, and 12 months later can be drawn starting from today's position. The displacement vector between the 0 month and 3 month positions can be calculated by taking the square root of the sum of the squares of the radius. The magnitude of the displacement vector for 3, 6, and 9 months can be calculated by adding
  • #1
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Homework Statement


The Earth moves around the Sun in a circle of radius 1.50 X 10^11 m at approximately constant speed. Taking today's position of the Earth as origin, draw a diagram showing the position vector 3 months, 6 months, 9 months, and 12 months later. Draw the displacement vector between the 0 month and 3 month positions and calculate the magnitude of the displacement vector for this 3-month interval.


Homework Equations


x= rcos[tex]\phi[/tex]
y= rsin[tex]\phi[/tex]
x^2 + y^2 = r^2



The Attempt at a Solution


my friend told me that the diagram is supposed to look like the sine function and that the amplitude= 1.5 X 10^11 m, but I still don't understand how to calculate the magnitude of the displacement vector.
 
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  • #2
Your friend told you wrong. The Earth goes around the Sun much like the tip of a clock hand goes around the face of the clock. If "today" the tip is at 12 o' clock (0 month position), it will be at the 3 o' clock position 3 months later, the 6 o'clock position 6 months later and so on. Can you draw the displacement vectors and calculate their magnitudes now?
 
  • #3
So do I just take the square root of ((1.50 X 10^11)^2 + (1.50 X 10^11)^2) to get the magnitude of the displacement vector?
 
  • #4
For 3 months, yes. What about 6 months?
 
  • #5
Between the 0 month and 6 month positions would I take (1.50 X 10^11) + (1.50 X 10^11)?
 
  • #6
Yup.
 

What is the magnitude of displacement vector?

The magnitude of displacement vector is the size or length of the displacement vector. It represents the distance between the initial and final positions of an object.

How is the magnitude of displacement vector calculated?

The magnitude of displacement vector can be calculated using the Pythagorean theorem where the magnitude is equal to the square root of the sum of the squares of the individual components of the vector.

What is the unit of measurement for magnitude of displacement vector?

The unit of measurement for magnitude of displacement vector depends on the units used for the individual components of the vector. For example, if the components are measured in meters, the magnitude will be in meters.

Can the magnitude of displacement vector be negative?

No, the magnitude of displacement vector is always a positive value as it represents a distance. However, the direction of the vector can be negative.

How does the magnitude of displacement vector differ from the distance traveled?

The magnitude of displacement vector represents only the straight-line distance between the initial and final positions. On the other hand, distance traveled takes into account the actual path taken by the object and can be longer than the magnitude of displacement vector.

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