Linear Speed of Object on Earth Surface at 62.0 Degrees Angle

In summary, the linear speed of an object on the surface of the Earth, relative to someone on the rotation axis, is 0.41 km/s or 4.10E+02 m/s. This can be found by multiplying the angular velocity of the Earth (7.29E-05 in radians/second) by the radius of the circle traced out by the object (6.37E+3 km), which can be determined using the formula r = R sin θ.
  • #1
Eleet
9
0
Assume the Earth is spherical. Relative to someone on the rotation axis, what is the linear speed of an object on the surface if the radius vector from the center of the Earth to the object makes an angle of 62.0 degrees with the axis of rotation. Radius of the Earth = 6.37·103 km

I know that the Earth is rotating, 7.29E-05 in radians/second.

Also, I think the formula to find linear speed for motion in a circle is the radius times the angular velocity. I get an angular velocity of 850480.10973937 r/s and I do not think this is right?
 
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  • #2
I'm not sure how you arrived at your answer, but the question is asking for the linear speed (not the angular velocity of object). The angular velocity of any object on the Earth is actually the same. To see this, note that the path each [stationary] object traces out in space is a circle; and, moreover, each object traces its circle in the same amount of time (namely 24 hours).

The linear speed of a body is equal to the product of the angular speed and radius of the circle it is tracing out. For the body related to the problem, the angular speed is known and the radius can be determined as r = R sin th. Now you can just plug in.
 
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  • #3
Eleet said:
Radius of the Earth = 6.37·103 km
OK, that's 6.37E+3 km.
I know that the Earth is rotating, 7.29E-05 in radians/second.
OK, that's the angular velocity.
Also, I think the formula to find linear speed for motion in a circle is the radius times the angular velocity.
Right. Find the radius of the circle traced out by that object.
I get an angular velocity of 850480.10973937 r/s and I do not think this is right?
Well... the angular velocity you already found. (That's the rotation of the Earth which you gave above.) Now find the linear speed (in m/s or km/s), using the idea you just supplied yourself. (I have no idea where the number you calculated came from. If you're still stuck, describe exactly how you calculated it.)
 
  • #4
So I multiply 7.29E-05 by sin 62 then multiply that by 6370km.

and I get 0.41 km/s or 4.10E+02 m/s

since I have this info do I find the objects acceleration by a= v2/r. r=6.37E+3 km
 
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  • #5
that does give the correct answer, but the reasoning is a little off. your multiplying the radius of the Earth by the sin 62 first in order to find the radius of the circle the object traces out. it would help to draw a picture to see this. you could take the origin to be the center of the earth, the y-axis to point to the north pole, and let the x-axis point toward the position on the equator that "lines up" with the object.
 

Related to Linear Speed of Object on Earth Surface at 62.0 Degrees Angle

1. What is the formula for calculating the linear speed of an object on the Earth's surface at a 62.0 degree angle?

The formula for calculating linear speed is v = d/t, where v is the linear speed in meters per second, d is the distance traveled in meters, and t is the time it takes to travel that distance in seconds. In this case, d would be the distance traveled along the Earth's surface, and t would be the time it takes to travel that distance at a 62.0 degree angle.

2. How is the linear speed affected by the angle of the object's motion on the Earth's surface?

The linear speed of an object on the Earth's surface is affected by the angle of motion because the distance traveled along the surface is different from the straight-line distance between the starting and ending points. The steeper the angle, the shorter the distance traveled and therefore the slower the linear speed. In the case of a 62.0 degree angle, the linear speed would be slower compared to a straight-line motion.

3. Is the linear speed on the Earth's surface at a 62.0 degree angle the same as the linear speed on a flat surface?

No, the linear speed on the Earth's surface at a 62.0 degree angle is not the same as the linear speed on a flat surface. This is because the Earth's surface is curved, and therefore the distance traveled along the surface is longer than the straight-line distance between the starting and ending points. As a result, the linear speed will be slower on the Earth's surface compared to a flat surface.

4. How can the linear speed of an object on the Earth's surface at a 62.0 degree angle be measured?

The linear speed of an object on the Earth's surface at a 62.0 degree angle can be measured using a stopwatch and a measuring tape. First, mark a starting point and an ending point on the Earth's surface at the desired angle. Then, have the object travel from the starting point to the ending point while timing it with the stopwatch. Measure the distance traveled using the measuring tape. Finally, plug the time and distance values into the formula v = d/t to calculate the linear speed.

5. How does the linear speed of an object on the Earth's surface at a 62.0 degree angle affect its motion?

The linear speed of an object on the Earth's surface at a 62.0 degree angle affects its motion by determining how far it will travel in a given amount of time. A slower linear speed means the object will travel a shorter distance in the same amount of time compared to a higher linear speed. This can affect the overall time it takes for the object to reach its destination and the amount of energy required for its motion.

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