Large stars can explode as they finish burning their nuclear fuel, causing a supernova. The explosion blows away the outer layers of the star. According to Newton’s third law, the forces that
push the outer layers away have reaction forces that are inwardly directed on the core of the star. These forces compress the core and can cause the core to undergo a gravitational collapse. The
gravitational forces keep pulling all the matter together tighter and tighter, crushing atoms out of existence. Under these extreme conditions, a proton and an electron can be squeezed together to
form a neutron. If the collapse is halted when the neutrons all come into contact with each other, the result is an object called a neutron star, an entire star consisting of solid nuclear matter. Many neutron stars rotate about their axis with a period of ≈ 1 s and, as they do so, send out a pulse of electromagnetic waves once a second. These stars were discovered in the 1960s and are called pulsars.
a. Consider a neutron star with a mass equal to the sun, a radius of 10 km, and a rotation period of 1.0 s. What is the speed of a point on the equator of the star?
b. What is g at the surface of this neutron star?
c. A stationary 1.0 kg mass has a weight on earth of 9.8 N. What would be its weight on the star?
d. How many revolutions per minute are made by a satellite orbiting 1.0 km above the surface?
e. What is the radius of a geosynchronous orbit?

Step by step solution

01

Part(a) Step1 : Given information

Radius of the star, R=10 km=10 x 10³m
Rotation Period, T=1 s
Mass of the star, m=1.98 x 10³⁰ kg

02

Part(a) Step2: Solution

Speed of the star is calculated by
$v=\frac{2\pi r}{T}$

Substitute values

$$\begin{gathered} v=\frac{2\pi \times (10\times {10}^{3}m)}{1s} \\ =6.3\times {10}^4\mathrm{m}/\mathrm{s} \end{gathered}$$

velocity is 6.3 x10⁴m/s

03

Part(b) Step1: Given information

Radius of the star, R=10 km=10 x 10³m
Rotation Period, T=1 s
Mass of the star, m=1.98 x 10³⁰ kg

04

Part(b) Step2: Explanation

We can calculate gravitational acceleration by

$g=\frac{GM}{r^2}$

Substitute the given values we get,

$$\begin{gathered} g=\frac{(6.67\times {10}^{-11}N.m^2/k{g}^2)\times (1.98\times {10}^{30}kg)}{{\left(10\times {10}^{3}m\right)}^2} \\ =1.3\times {10}^{12}\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Acceleration due to gravity is 1.2 x 10¹² m/s²

05

Part(c) Step1: Given information

Radius of the star, R=10 km=10 x 10³m
Rotation Period, T=1 s
Mass of the star, m=1.98 x 10³⁰ kg

mass of person is 1 kg.

06

Part(c) Step2: Explanation

Weight can be calculated as
W=m g
Substitute the values

W =1 kg x 1.3 x 10¹² m/s²

W=1.3 x 10¹² N

07

Part(d) Step1: Given information

Radius of the star, R=10 km=10 x 10³m
Rotation Period, T=1 s
Mass of the star, m=1.98 x 10³⁰ kg

mass of person is 1 kg.

radius of orbit = 1 km = 1000 m

08

Part(d) Step2: Explanation

In the given condition

Centripetal acceleration= Gravitational acceleration

so,

$$\begin{gathered} \frac{v^2}{r}=\frac{GM}{r^2} \\ {v}^2=\frac{(6.67\times {10}^{-11}N/k{g}^2.m^2)\times (1.98\times {10}^{30}kg)}{{(1\times {10}^{3}m)}^3} \\ v=3.69\times {10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Now calculate number of revolution

$$\begin{gathered} v=\frac{2\pi rN}{60} \\ 3.69\times {10}^{8}m/s=\frac{2\pi \times 10\times {10}^{3}m}{60}\times N \\ N=35.2\times {10}^4\mathrm{rpm} \end{gathered}$$

No of revolution is 35.2 x 10⁴ rpm.

09

Part(e) Step1: Given information

Radius of the star, R=10 km=10 x 10³m
Rotation Period, T=1 s
Mass of the star, m=1.98 x 10³⁰ kg

10

Part(e) Step2: Explanation

Calculate radius of the orbit as

$$\begin{gathered} T=\frac{2\pi r}{\sqrt{\frac{GM}{r}}} \\ \mathrm{Substitute}\mathrm{values}\mathrm{and}\mathrm{solve}\mathrm{for}\mathrm{r} \\ 1=\frac{2\pi r}{\sqrt{\frac{(6.67\times {10}^{-11}N/k{g}^2.m^2)\times (1.98\times {10}^{30}m/s^2)}{r}}} \\ r=1.5\times {10}^6\mathrm{m} \end{gathered}$$

Radius is 1.5 x 10⁶ m.

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