Question

Question: (a) If the legendary apple of Newton could be released from rest at a height of 2 m from the surface of a neutron star with a mass 1.5 times that of our Sun and a radius of20 km, what would be the apple’s speed when it reached the surface of the star? (b) If the apple could rest on the surface of the star, what would be the approximate difference between the gravitational acceleration at the top and at the bottom of the apple? (Choose a reasonable size for an apple; the answer indicates that an apple would never survive near a neutron star.)

Short answer

Answer:

The speed of the apple when it reaches the surface of the neutron star$v=1.4\times {10}^6\frac{m}{s}.$

The approximate difference between the top and bottom of an apple $a_g=3\times {10}^6\frac{m}{s^2}.$

Step-by-step solution

Identification of given data

The height of the apple from the surface h =2m,

The mass of the neutron star is

$$\begin{gathered} M=1.5\times M_{sun} \\ =2.98\times {10}^{30}kg \end{gathered}$$

The radius of neutron star is

$$\begin{gathered} r=20 km \\ =2.0\times {10}^{4}m \end{gathered}$$

Significance of conservation of energy

According to the rule of conservation of energy, energy can only be transformed from one form of energy to another and cannot be created or destroyed.

We can find the speed of the apple on the surface of the neutron star using the law of conservation of energy for the object. Then we can find the gravitational acceleration on the surface and at height h. From this, we can find the approximate difference between the top and bottom of the apple.

Formula:

$$\begin{gathered} \frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}^2-\frac{GMm}{r} \\ a=\frac{GM}{r^2} \end{gathered}$$

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

a is the gravitational acceleration

(a) Determining the speed of the apple when it reaches the surface of the neutron star

To calculate the velocity at the surface of the neutron star, we can use the equation of conservation of energy.

We have

$\frac{1}{2}m{v}_1^2-\frac{GMm}{r_1}=\frac{1}{2}m{v}^2-\frac{GMm}{r}$

Here, we can cancel out on both sides of the equation. Then substituting we get,

$$\begin{gathered} 0-\frac{GM}{r_1}=\frac{1}{2}{v}^2-\frac{GM}{r} \\ \frac{1}{2}{v}^2=\frac{GM}{r}-\frac{GM}{r_1} \\ v=\sqrt{\frac{2GM\left(r_1-r\right)}{r_{1}r}} \\ =\sqrt{\frac{2\left(6.67\times {10}^{-11}\frac{N.m^2}{k{g}^2}\right)\left(2.98\times {10}^{30} \text{kg}\right)\left(20002 \text{m}-20000 \text{m}\right)}{20002 \text{m}\left(20000 \text{m}\right)}} \\ =1.4\times {10}^6\frac{\text{m}}{\text{s}} \end{gathered}$$

Therefore, the speed of the apple when it reaches the surface of the neutron star is
$1.4\times {10}^6\frac{m}{s}.$

(b) Determining the approximate difference between the top and bottom of an apple

We have the equation for gravitational acceleration as

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

Using this equation, we can get the gravitational acceleration at the bottom and at the top of the apple. Let us assume the apple to be of length 6 cm = 0.06 m.

$$\begin{gathered} a_b=\frac{GM}{r^2} \\ =\frac{\left(6.67\times {10}^{-11}\frac{N.m^2}{k{g}^2}\right)\left(2.98\times {10}^{30} \text{kg}\right)}{{\left(2.0\times {10}^4 \text{m}\right)}^2} \\ =496915000000 \frac{m}{s^2} \end{gathered}$$

In a similar way, for the top of apple,

$$\begin{gathered} R=r+h \\ =20000.06 \text{m} \end{gathered}$$

So.

$a_t=496912000000 \frac{m}{s^2}$

So, the difference in acceleration is

$$\begin{gathered} a_g=a_b-a_t \\ =3000000 \frac{m}{s^2} \\ {a}_g=3\times {10}^6\frac{m}{s^2} \end{gathered}$$

Therefore, the approximate difference between the top and bottom of the apple is

$3\times {10}^6\frac{m}{s^2}$