Choose the correct alternative :

**(a)** Acceleration due to gravity increases/decreases with increasing altitude.

**(b)** Acceleration due to gravity increases/decreases with increasing depth (assume the earth to be a sphere of uniform density).

**(c)** Acceleration due to gravity is independent of the mass of the earth/mass of the body.

**(d)** The formula –G M m(\( \dfrac{1}{r_2}-\dfrac{1}{r_1}\)) is more/less accurate than the formula mg(r_{2} – r_{1}) for the difference of potential energy between two points r_{2} and r_{1} distance away from the centre of the earth.

Asked by Pragya Singh | 1 year ago | 77

**(a)** Acceleration due to gravity **decreases** with increasing altitude.

**(b)** Acceleration due to gravity **decreases** with increasing depth (assume the earth to be a sphere of uniform density).

**(c)** Acceleration due to gravity is independent of the mass of the **mass of the body**.

**(d) **The formula –G M m(\( \dfrac{1}{r_2}\) – \( \dfrac{1}{r_1}\)) ** more **accurate than the formula mg(r_{2} – r_{1}) for the difference of potential energy between two points r_{2} and r_{1} distance away from the centre of the earth.

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 = 6.4×10^{23} kg; radius of mars = 3395 km; G = 6.67×10^{-11} N m^{2} kg^{–2}.

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 = 2×1030 kg; mass of mars = 6.4×1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 ×108 km; G = 6.67×10-11 N m2kg–2.

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 = 2×10^{30} kg).

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 = 6.0×10^{24} kg, radius = 6400 km

Two heavy spheres each of mass 100 kg and radius 0.10 m are placed 1.0 m apart on a horizontal table. What is the gravitational force and potential at the midpoint of the line joining the centres of the spheres? Is an object placed at that point in equilibrium? If so, is the equilibrium stable or unstable?