Assertion (A): Comet tail points away from the sun.
Reason (R): Solar radiation vapourise the volatile materials within the comet.
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.
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
By conservation of energy, \(v_{\infty} = \sqrt{v^2 - v_{\text{esc}}^2} = \sqrt{(2v_{\text{esc}})^2 - v_{\text{esc}}^2} = v_{\text{esc}}\sqrt{3} = 11.2\sqrt{3} \text{km/s}\).
An artificial satellite is moving in a circular orbit of radius \(r\) around a planet. Total energy of satellite is \(E\). Energy required to move this satellite into a new orbit of radius \(2r\), is
The total energy in orbit is \(E = -\frac{GMm}{2r}\). In the new orbit of radius \(2r\), total energy is \(E' = -\frac{GMm}{4r} = \frac{E}{2}\). The required energy is \(\Delta E = E' - E = \frac{E}{2} - E = -\frac{E}{2}\).
The factors on which the escape speed from earth depend, is/are
Escape velocity is given by \(v_e = \sqrt{\frac{2GM}{R+h}}\). It depends on the mass of the earth \(M\) and the height of projection \(h\), but is independent of the mass of the projected object.
Two bodies each of mass \( 1\text{ kg}\) are placed \( 2\text{ m}\) apart. Gravitational potential energy of the system is (Assume potential energy to be zero at infinity)
The gravitational potential energy of a two-body system is given by \( U = -\frac{G m_1 m_2}{r} \). Substituting \( m_1 = m_2 = 1\text{ kg} \) and \( r = 2\text{ m} \), we get \( U = -\frac{G}{2} \).
Three equal masses of \(3\text{ kg}\) each are fixed at the vertices of an equilateral triangle ABC. The gravitational force acting on mass \(2\text{ kg}\) placed at the centroid of triangle is
Due to perfect symmetric distribution, the three gravitational pull forces on the mass at the centroid are equal in magnitude and separated by \(120^circ\), resulting in a net vector sum of zero.
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 \(R\)]
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.