A pulsar (an imploding star) rotates at an incredible rate of 30 times a second. If its diameter is 12 km, find the speed of a point on its equator. Give your answer correct to 3 significant figures.



Answer :

Certainly! Let's break down the problem step by step to find the speed of a point on the equator of the pulsar.

1. Understand the given information:
- The pulsar rotates 30 times per second.
- The diameter of the pulsar is 12 km.

2. Find the radius:
- The radius is half of the diameter. Thus, the radius is:
[tex]\[ \text{Radius} = \frac{\text{Diameter}}{2} = \frac{12 \text{ km}}{2} = 6 \text{ km} \][/tex]

3. Calculate the circumference of the circle (equator) of the pulsar:
- The circumference of a circle is given by the formula:
[tex]\[ \text{Circumference} = 2 \pi \times \text{Radius} \][/tex]
- Substituting the radius we found:
[tex]\[ \text{Circumference} = 2 \pi \times 6 \text{ km} = 37.69911184307752 \text{ km} \][/tex]

4. Determine the linear speed:
- Linear speed can be found by multiplying the circumference by the rotation rate:
[tex]\[ \text{Speed} = \text{Circumference} \times \text{Rotation rate} = 37.69911184307752 \text{ km} \times 30 \text{ rotations per second} \][/tex]
- Performing this calculation:
[tex]\[ \text{Speed} = 1130.9733552923256 \text{ km/s} \][/tex]

5. Round the result to 3 significant figures:
- Rounding 1130.9733552923256 to 3 significant figures:
[tex]\[ \text{Rounded speed} = 1130.973 \text{ km/s} \][/tex]

Therefore, the speed of a point on the equator of the pulsar, correct to 3 significant figures, is [tex]\( 1130.973 \text{ km/s} \)[/tex].

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