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Q8.23:
A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity ? (mass of the sun = $2 \times 10^{30}$ kg).
Solution :
Step 1: Given Variables and Initial Setup
To determine if an object remains stuck to the surface, we must compare the gravitational acceleration ($g'$) at the equator with the centripetal acceleration ($a_c$) required for circular motion due to the rotation of the star.
- Mass of the star ($M$): $2.5 \times M_{sun} = 2.5 \times (2 \times 10^{30} \text{ kg}) = 5.0 \times 10^{30} \text{ kg}$
- Radius of the star ($R$): $12 \text{ km} = 1.2 \times 10^4 \text{ m}$
- Frequency of rotation ($f$): $1.2 \text{ rev/s}$
- Gravitational constant ($G$): $6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$
Step 2: Calculating Gravitational Acceleration ($g'$)
The acceleration due to gravity at the surface of the star is given by the formula:
$g' = \frac{GM}{R^2}$
Substituting the values:
$g' = \frac{(6.67 \times 10^{-11}) \times (5.0 \times 10^{30})}{(1.2 \times 10^4)^2}$
$g' = \frac{33.35 \times 10^{19}}{1.44 \times 10^8} \approx 2.316 \times 10^{12} \text{ m/s}^2$ [Per Newton's Law of Universal Gravitation]
Step 3: Calculating Centripetal Acceleration ($a_c$)
The centripetal acceleration required to maintain an object on the surface is $a_c = \omega^2 R$, where $\omega = 2\pi f$.
$\omega = 2 \times 3.14159 \times 1.2 \approx 7.54 \text{ rad/s}$
$a_c = (7.54)^2 \times (1.2 \times 10^4)$
$a_c \approx 56.85 \times 1.2 \times 10^4 \approx 6.82 \times 10^5 \text{ m/s}^2$
Step 4: Comparison and Conclusion
An object will remain stuck to the surface if the gravitational pull exceeds the required centripetal acceleration ($g' > a_c$).
| Parameter | Magnitude ($\text{m/s}^2$) |
|---|---|
| Gravitational Acceleration ($g'$) | $2.316 \times 10^{12}$ |
| Centripetal Acceleration ($a_c$) | $6.82 \times 10^5$ |
Since $g' \gg a_c$ (by several orders of magnitude), the gravitational force is significantly stronger than the centrifugal effect of the rotation.
Final Solution: Yes, the object will remain stuck to the surface because the gravitational acceleration ($2.316 \times 10^{12} \text{ m/s}^2$) is vastly greater than the acceleration required to fling it off the surface ($6.82 \times 10^5 \text{ m/s}^2$).
More Questions from Class 11 Physics Gravitation Additional Exercises
- Q8.22: As you have learnt in the text, a geostationary satellite orbits the earth at a height of nearly 36,000 km from the surface of the earth. What is the potential due to earth’s gravity at the site of this satellite ? (Take the potential energy at infinity to be zero). Mass of the earth = $6.0 \times 10^{24}$ kg, radius = 6400 km.
- Q8.24: A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system ? Mass of the space ship = 1000 kg; mass of the sun = $2 \times 10^{30}$ kg; mass of mars = $6.4 \times 10^{23}$ kg; radius of mars = 3395 km; radius of the orbit of mars = $2.28 \times 10^8$ km; G = $6.67 \times 10^{-11}$ N m$^2$ kg$^{–2}$.
- Q8.25: A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s$^{–1}$. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it ? Mass of mars = $6.4 \times 10^{23}$ kg; radius of mars = 3395 km; G = $6.67 \times 10^{-11}$ N m$^2$ kg$^{–2}$.
CBSE Solutions for Class 11 Physics Gravitation
Chapters in CBSE - Class 11 Physics
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