Question

Question: Consider a pulsar, a collapsed star of extremely high density, with a mass equal to that of the Sun $\left(1.98\times {10}^{30}kg\right)$, a radiusRof only 12 km , and a rotational period T of 0.041s . By what percentage does the free-fall acceleration gdiffer from the gravitational acceleration a_gat the equator of this spherical star?

Short answer

Answer:

The percentage by which the free fall acceleration g differs from the gravitational acceleration a_g at the equator of the spherical star is 0.031%

Step-by-step solution

Identification of given data

The mass of the pulsar is $M=1.98\times {10}^{30}kg$

The radius of the pulsar is

$$\begin{gathered} R=12 km \\ =12\times {10}^{3}m \end{gathered}$$

The rotational period of the pulsar is $T=0.041 s$

Significance of gravitational acceleration

The term "gravitational acceleration" refers to the acceleration that an object experiences as a result of the force of gravity acting on it.

Using the expression for the effect of rotational motion of earth on the gravitational acceleration of earth, we can find the percentage difference between the free fall acceleration from the gravitational acceleration at the equator of the spherical star.

Formula:

$$\begin{gathered} a_g=\frac{GM}{R^2} \\ g=a_g-{\omega }^{2}R \end{gathered}$$

Where, G is the gravitational constant ( $6.67\times {10}^{-11} \frac{N.m^2}{k{g}^2}$)

Determining how much percentage does the free-fall acceleration g  differ from the gravitational acceleration ag  at the equator of this spherical star

The percentage of the free fall acceleration g differs from the gravitational acceleration a_g at the equator of the spherical star:

The gravitational acceleration on the surface of the pulsar is $a_g=\frac{GM}{R^2}$

According to the expression for the effect of rotational motion of earth on the gravitational acceleration,

$$\begin{gathered} g=a_g-{\omega }^{2}R \\ {a}_g-g={\omega }^{2}R \\ \frac{a_g-g}{a_g}=\frac{{\omega }^{2}R}{a_g} \end{gathered}$$

$$\begin{gathered} \omega =\frac{2\pi }{T} \\ \frac{a_g-g}{a_g}=\frac{{\left(\frac{2\pi }{T}\right)}^{2}R}{\frac{GM}{R^2}} \\ =\frac{{\left(\frac{2\pi }{T}\right)}^2{R}^3}{GM} \\ \frac{a_g-g}{a_g}=\frac{{\left(\frac{2\times 3.142}{0.041\text{s}}\right)}^2{\left(12\times {10}^3\right)}^3}{\left(6.67\times {10}^{-11}\frac{{\text{Nm}}^{\text{2}}}{{\text{kg}}^{\text{2}}}\right)\times \left(1.98\times {10}^{30}\text{kg}\right)} \\ =3.1\times {10}^{-4} \\ =0.031\% \end{gathered}$$

The percentage by which the free fall acceleration g differs from the gravitational acceleration a_g at the equator of the spherical star is 0.031%