The Crab pulsar (m=2.00x10^30 kg) is a neutron star located in the Crab Nebula. The rotation rate of the Crab pulsar is currently about 30.0 rotations per second, or 60.0pi rad/s. The rotation rate of the pulsar, however, is decreasing; each year, the rotation period increases by 1.00x10^-5 s. Justify the following statement: The loss in rotational energy of the pulsar to 1.0010^5 times the power output of the Sun. ( The total power radiated by the Sun is about 4.0010^26 W.)

Question

The Crab pulsar (m=2.00x10^30 kg) is a neutron star located in the Crab Nebula. The rotation rate of the Crab pulsar is currently about 30.0 rotations per second, or 60.0pi rad/s. The rotation rate of the pulsar, however, is decreasing; each year, the rotation period increases by 1.00x10^-5 s. Justify the following statement: The loss in rotational energy of the pulsar to 1.0010^5 times the power output of the Sun. ( The total power radiated by the Sun is about 4.0010^26 W.)

P/Ps=_________

Answer 1

rotational energy loss= 1/2 I (wi^2-wf^2)

= 1/2 I (wi+wf)(wi-wf)
now the moment of inertia for a solid sphere is 2/5 mr^2

= 1/2 * 2/5*2E30 * (wi*r-wf*r)(wi*r+wi*r)

now period= 1/freq=1/30
so w=2PI/period

wi=2PI/30
wf=2PI/(30-1E-5)

so, finally, (wi^2-wf^2)=
= (2PI)^2 (1/30^2 -1/(30-1E-5)^2
= (2PI)^2 ( 900-60E-5-1E-10 -900)/(30^3*(30-1E-5)^2

= appx (2PI)^2 ( 1/15 * 1E-5)

then, power loss is appx
1/5 *2E30*(2PI)^2 (1/15 E-5)
= 2.6E24
so recheck my work.

Answer 2

To justify the given statement, we need to calculate the loss in rotational energy of the pulsar and compare it to 1.00x10^5 times the power output of the sun.

Let's break down the problem step by step:
1. Determine the change in rotation period per year.
2. Calculate the change in rotational energy per year.
3. Compare the change in rotational energy to the power output of the sun.

Step 1: Change in rotation period per year
The rotation period of the pulsar increases by 1.00x10^-5 seconds per year.
Since the initial rotation rate is 60.0π rad/s, the change in rotation rate per year can be calculated as follows:
Δω = 2π / (initial rotation period + change in rotation period per year) - 60.0π

Step 2: Change in rotational energy per year
The change in rotational energy per year (ΔE) can be calculated using the formula:
ΔE = (1/2)I(ωi^2 - ωf^2)

Since we know the initial rotation rate (ωi), the final rotation rate (ωf), and the mass of the pulsar (m), we can calculate the moment of inertia (I) using the formula:
I = (2/5)mr^2

The radius of the pulsar is not given, but we can assume it to be a typical neutron star radius, which is around 10 km (1.0 x 10^4 m).

Substituting the values, we can calculate ΔE.

Step 3: Compare change in rotational energy to the power output of the sun
To compare the change in rotational energy to the power output of the sun, we need to calculate the power output of the pulsar per year.

The power output (P) is given by the formula:
P = ΔE / t (where t is the time in seconds)

Finally, we can compare P to 1.00x10^5 times the power output of the sun (4.00x10^26 W) using the formula:
P / (1.00x10^5 * Power output of the sun)

By evaluating this expression, we can determine the ratio P/Ps.

Note: The units and values provided in the problem statement need to be used consistently throughout the calculations to obtain accurate results.