Question

Info given:
Rotation period: \[T=33m/s\]
\[M=3\cdot10^{30}\]
\[R=\cdot10^{4}\]

A pulsar radiates energy:
\[P=6\cdot10^{31}W\]. How much longer will the pulsars period be after 24h?

Answer 1

\[K=\dfrac{1}{2}I\cdot\omega^2\]
Where
\[I=1.2\cdot10^{38} kg\cdot m^2 \]\[\omega=(\dfrac{33 \dfrac{m}{s}}{10^4m})=0.0033 \dfrac{rad}{s}\]

Answer 2

\[K_{pulsar1}=\dfrac{1}{2}\cdot1.2\cdot10^{38}\cdot0.0033=6.53\cdot10^{32}J\]and
\[K_{pulsar2}=\dfrac{6 \cdot 10^{21}}{24\cdot60\cdot60}=6.94\cdot10^{26}J\]
The lost energy over 24h:
\[K_{pulsar3}=K_{pulsar1}-K_{pulsar2}\]<=>
\[K_{pulsar3}=6.53\cdot10^{32}J-6.94\cdot10^{26}J\]<=>
\[K_{pulsar3}=K_{pulsar1}-K_{pulsar2}\]<=>
\[K_{pulsar3}=6.53\cdot10^{32}J\]

Answer 3

The new period should then be \[K=\dfrac{1}{2}⋅I⋅ω^2\]<=>\[K_{pulsar3}=\dfrac{1}{2}⋅1.2\cdot10^{38}⋅ω^2\]
Where i solve for ω to get the new period:
\[ω=0.0033 \dfrac{rad}{s}\]

Answer 4

My problem is, that it's the same value as the one i started with ...

Answer 5

is this from edx? :D

Answer 6

ur omega is wrong

T = 33ms.. not m PER SECOND as u have written.. its time period.. not linear speed!

Answer 7

edx ?

Answer 8

Oh you mean milliseconds ?

Answer 9

yeha.. milliseconds..

MIT X.. online course? or you are actually from MIT ?

Answer 10

No im not from MIT :) im a first year student from a danish university.

Answer 11

I found the problem actually, it was a calculation mistake :)

Answer 12

\[\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{33ms}\]
would make more sense ye ?

Answer 13

Solved it x)!