Gravitation
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See all →- A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is : Answer: 2. 27 N Key idea: weight at height h varies as 1/(R+h)² . F=48(R/R+R/3)²=48×(3/4)²=48×frac916=27\ mathrm N 1
- The radius of Martian orbit around the Sun is about 4 times the radius of the orbit of Mercury. The Martian year is 687 Earth days. Then which of the following is the length of 1 year on Mercury ? Answer: 1. 88 earth days Key idea: Kepler's third law, T²∝ r³ . fracT_MarsT_Mercury=(fracr_Marsr_Mercury)^(3/2)=4^(3/2)=8 T_Mercury=687/8≈86... 1
- The minimum energy required to launch a satellite of mass m from the surface of earth of mass M and radius R in a circular orbit at an altitude of 2R from the surface of the earth is: Answer: 2. 5GmM/6R Key idea: The minimum energy is the difference between the total energy in the orbit and the total energy at the surface.... 1
- The mass of a planet is 1/10th that of the earth and its diameter is half that of the earth. The acceleration due to gravity on that planet is : Answer: 1. 3.92 m s ⁻² Key idea: g=GM/R² . For the planet M_p=M/10 and R_p=R/2 , so g_p=G(M/10)/(R/2)²=4/10 g=0.4×9.8=3.92\ m s⁻² Option 1. 1
- Two bodies of mass m and 9m are placed at a distance R. The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be (G = gravitational constant) : Answer: 4. -16Gm/R Find the point where the net field is zero, then add the potentials of both masses there. Let the point be at distance x... 1
- A satellite is orbiting just above the surface of the earth with period T. If d is the density of the earth and G is the universal constant of gravitation, the quantity 3π/Gd represents : Answer: 3. T² For a satellite just above the surface, r=R and T=2π√(R³/GM) , so T²=4π²R³/GM . With M=4/3π R³d : T²=4π²R³/G·4/3π R³d=3π/Gd .... 1