To solve this problem, we need to find both the speed and the magnitude of the acceleration of a particle at the neutron star's equator. Let's break it down step by step.

1. Speed of a particle at the equator

The speed $v$ of a particle at the star's equator due to the rotation can be found using the formula:

$v = \frac{2 \pi R}{T}$

Where:

• $R$ is the radius of the neutron star,
• $T$ is the period of rotation (time for one complete rotation).

Given:

• Radius $R = 21 \, \text{km} = 21,000 \, \text{m}$,
• Period $T = 0.98 \, \text{seconds}$.

Now, plug in the values:

$v = \frac{2 \pi \times 21,000}{0.98}$

Let me calculate this.

2. Magnitude of the acceleration vector

The acceleration is centripetal acceleration, which can be calculated using:

$a = \frac{v^2}{R}$

We'll use the velocity $v$ calculated in the first part to find the acceleration.

Let me calculate both the speed and the acceleration.1. The speed of a particle at the neutron star's equator is approximately $134,640 \, \text{m/s}$.

2. The magnitude of the acceleration vector is approximately $863,231 \, \text{m/s}^2$.

Would you like further details on these calculations or any specific concepts clarified?

Here are 5 related questions:

1. How is the concept of centripetal force related to the acceleration in circular motion?
2. How does the size of the neutron star affect the speed and acceleration at the equator?
3. What would happen to the particle’s speed if the rotation period of the neutron star decreased?
4. How would the mass of the neutron star affect the gravitational pull at its surface?
5. How does relativistic speed (close to the speed of light) affect objects near neutron stars?

Tip: Centripetal acceleration is always directed towards the center of the circular path, keeping the object moving in a circle.