Assertion (A): Escape velocity of a satellite is greater than its orbital velocity.
Reason (R): Orbit of a satellite is within the gravitational field of planet whereas escaping is beyond the gravitational field of planet.
1. Both (A) & (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) & (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
Assertion (A) is true: escape velocity \(V_e = \sqrt{2GM/r}\) is \(\sqrt{2}\) times orbital velocity \(V_o = \sqrt{GM/r}\) for a circular orbit.
Reason (R) is false because the gravitational field extends infinitely. Escaping means overcoming the gravitational potential, not leaving the field.
Assertion (A): Comet tail points away from the sun.
Reason (R): Solar radiation vapourise the volatile materials within the comet.
1. Both (A) \(&\) (R) are true and the (R) is the correct explanation of the (A)
2. Both (A) \(&\) (R) are true but the (R) is not the correct explanation of the (A)
3. (A) is true but (R) is false
4. Both (A) and (R) are false
View Answer
The comet tail always points away from the Sun due to solar wind and radiation pressure pushing on the sublimated material. Solar radiation does vaporize volatile materials, forming the tail, but this vaporization itself doesn't fully explain the direction. Hence, both statements are true, but R is not the correct explanation of A.
If \(R\) is the radius of the earth and \(g\) is the acceleration due to gravity on the earth surface. Then the mean density of the earth will be
1. \(\frac{3g}{4\pi RG}\)
2. \(\frac{4\pi G}{3gR}\)
3. \(\frac{\pi RG}{12g}\)
4. \(\frac{3\pi R}{4gG}\)
View Answer
We know \(g = \frac{G M}{R^2} = \frac{G}{R^2} \left(\frac{4}{3}\pi R^3 \rho\right) = \frac{4}{3}\pi RG\rho\). Solving for density, \(\rho = \frac{3g}{4\pi RG}\).
The escape velocity of a body on the earth surface is 11.2 km/s. If the same body is projected upward with velocity 22.4 km/s, the velocity of this body at infinite distance from the centre of the earth will be
1. 11.2 km/s
2. \(11.2\sqrt{3} \text{km/s}\)
3. \(11.2\sqrt{2} \text{km/s}\)
4. Zero
View Answer
By conservation of energy, \(v_{\infty} = \sqrt{v^2 - v_{\text{esc}}^2} = \sqrt{(2v_{\text{esc}})^2 - v_{\text{esc}}^2} = v_{\text{esc}}\sqrt{3} = 11.2\sqrt{3} \text{km/s}\).
Which of the following statements are true about acceleration due to gravity (g)? (where symbols have their usual meaning)?
1. g will not change at poles if earth stops rotating on its axis
2. The value of acceleration due to gravity decreases as we move away from surface of earth
3. The value of acceleration due to gravity decreases as we move from surface towards centre of earth
4. All of these
View Answer
All statements are correct. Rotation does not affect gravity at the poles, and gravity decreases both above and below the surface of the Earth.
A rocket is fired vertically with a speed half of the escape speed from the earth’s surface. How far from the earth does the rocket go before returning to the earth? [Given mean radius of earth is \(R\)]
1. R/3 from earth’s centre
2. 4R/3 from the earth’s surface
3. 4R/3 from the earth’s centre
4. 7R/3 from the earth’s surface
View Answer
Using conservation of energy: \(-\frac{GMm}{R} + \frac{1}{2}m v^2 = -\frac{GMm}{r}\). Putting \(v = \frac{v_e}{2} = \sqrt{\frac{GM}{2R}}\) yields \(r = \frac{4R}{3}\), which is the distance from the earth's centre.