Question

A typical neutron star may have a mass equal to that of the Sun but a radius of only 10 Km . (a) What is the gravitational acceleration at the surface of such a star? (b) How fast would an object is moving if it fell from rest through a distance of1.0 mon such a star? (Assume the star does not rotate.)

Short answer

a. The gravitational acceleration at the surface of the star is $1.33\times {10}^{12}m/s^2$

b. The speed of object is $1.62\times {10}^{6}m/s$

Step-by-step solution

Identification of given data

Mass of star equal to mass of sun

Radius of star is $r=10.0km or 10000 m$

Distance of object is 1.0 m

Significance of Newton’s law of universal gravitation

Every particle in the universe is attracted to every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance, according to Newton's Law of Universal Gravitation.

With the mass and radius of a star, we can find its gravitational acceleration. Using the law of energy conservation, we can find its velocity.

Formula:

$a_g=\frac{GM}{r^2}$ …(i)

$K=\frac{1}{2}\times M\times V^2$ …(ii)

$U=-\frac{GMm}{r}$ …(iii)

Where, a_g is gravitational acceleration

G is the gravitational constant ( $6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}$)

M is the mass of star = mass of sun(role="math" $2.0\times {10}^{30} \text{kg}$)

K is the kinetic energy

U is the potential energy

V is the speed of object

(a) Determining the gravitational acceleration at the surface of the star 

From equation (i)

$$\begin{gathered} a_g=\frac{GM}{r^2} \\ =\frac{\left(6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}\right)\left(2.0\times {10}^{30} \text{kg}\right)}{{\left(10000 \text{m}\right)}^2} \\ =1.33\times {10}^{12} {\text{m/s}}^{\text{2}} \end{gathered}$$

(b) Determining the speed of object

By using energy conservation law,

$K{E}_{initial}+P{E}_{initial}=K{E}_{final}+P{E}_{final}$

From equation (ii) and equation (iii)

$$\begin{gathered} 0+\frac{GMm}{r_0}=\frac{1}{2}m{v}^2+\frac{GMm}{r} \\ \frac{1}{2}{v}^2=\frac{GM}{r}-\frac{GM}{r_0} \\ {v}^2=2GM\left(\frac{1}{r}-\frac{1}{r_0}\right) \\ v=\sqrt{2GM\left(\frac{1}{r}-\frac{1}{r_0}\right)} \end{gathered}$$

After plugging the values, we get $r_0=10001 \text{m}$

$$\begin{gathered} v=\sqrt{2\left(6.67\times {10}^{-11} {\text{m}}^{\text{3}}\text{/kg} ·{\text{s}}^{\text{2}}\right)\left(2.0\times {10}^{30} \text{kg}\right)\left(\frac{\text{1}}{10000 \text{m}}-\frac{1}{10001 \text{m}}\right)} \\ v=1.62\times {10}^6 \text{m/s} \end{gathered}$$

Hence, the speed of object is $1.62\times {10}^6 \text{m/s}$