Gravitation
- 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
- A body of mass 60 g experiences a gravitational force of 3.0 N, when placed at a particular point. The magnitude of the gravitational field intensity at that point is : Answer: 3. 50 N/kg Gravitational field intensity is force per unit mass: E=F/m . m=60\ g=0.06\ kg⇒ E=3.0/0.06=50\ N/kg . Option (3). Using 6... 1
- A particle of mass 'm' is projected with a velocity v = kV_e (k < 1) from the surface of the earth. (V_e = escape velocity)The maximum height above the surface reached by the particle is : Answer: 4. Rk²/1-k² Key idea: conserve energy: 1/2mv²-dfracGMmR=-GMm/R+h , with V_e²=dfrac2GMR . 1/2m k²dfrac2GMR-dfracGMmR=-GMm/R+h⇒dfracGM... 1
- The escape velocity from the Earth's surface is v. The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is : Answer: 4. 4v Key idea: v_e=√(2GM/R) and M=4/3π R³ρ , so v_e∝ Rsqrtρ . Same density and radius four times as large gives v_e'=4v . Option 4. 1
- A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth ? Answer: 2. 32 N Gravitational force varies as g , and g at height h is g(R/R+h)² . With h=R/2 : (R/1.5R)²=4/9 . F=72×4/9=32\ N Answer: optio... 1
- A body weighs 200 N on the surface of the earth. How much will it weigh half way down to the centre of the earth ? Answer: 4. 100 N Inside the earth at depth d , g'=g(1-d/R) . Halfway to the centre means d=R/2 , so g'=g/2 . Weight =200/2=100 N. 1
- The work done to raise a mass m from the surface of the earth to a height h, which is equal to the radius of the earth, is : Answer: 3. 1/2mgR Work done = change in gravitational potential energy. W=-GMm/R+h+GMm/R . With h=R : W=GMm(1/R-dfrac12R)=GMm/2R . Using GM=... 1
- The kinetic energies of a planet in an elliptical orbit about the Sun, at positions A, B and C are K_A, K_B and K_C, respectively. AC is the major axis and SB is perpendicular to AC at the position of the Sun S as shown in the figure. Then Answer: 3. K_A>K_B>K_C By conservation of angular momentum (Kepler's second law), a planet moves faster when it is closer to the Sun, so its... 1
- If the mass of the Sun were ten times smaller and the universal gravitational constant were ten times larger in magnitude, which of the following is not correct ? Answer: 4. 'g' on the Earth will not change. Check each statement with M_sun→ M/10 and G→10G . The Earth's g=GM_e/R_e² becomes 10 times larg... 1
- The acceleration due to gravity at a height 1 km above the earth is the same as at a depth d below the surface of earth. Then : Answer: 3. d = 2 km Key idea: at height h (small): g'=g(1-2h/R) . At depth d : g'=g(1-d/R) . Equating: 2h/R=d/R⇒ d=2h=2 km. 1
- Two astronauts are floating in gravitational free space after having lost contact with their spaceship. The two will : Answer: 1. move towards each other. Key idea: the only force between the two astronauts is their mutual gravitational attraction. In free sp... 1
- Starting from the centre of the earth having radius R, the variation of g (acceleration due to gravity) is shown by Answer: 1. graph (1) Inside the earth g is proportional to the distance from the centre; outside it falls as 1/r² . For r For r>R : g=GM/r²... 1
- A satellite of mass m is orbiting the earth (of radius R) at a height h from its surface. The total energy of the satellite in terms of g_0, the value of acceleration due to gravity at the earth's surface, is Answer: 1. -mg₀R²/2(R+h) For a satellite in orbit, total energy is half of its (negative) potential energy: E=-GMm/2(R+h) . Use g₀=GM/R² , s... 1
- The ratio of escape velocity at earth (v_e) to the escape velocity at a planet (v_p) whose radius and mean density are twice as that of earth is : Answer: 2. 1 : 2√(2) Key idea: v_e=√(2GM/R)=R√(8π Gρ/3) , so v_e∝ Rsqrtρ . Planet: radius 2R , density 2ρ , so v_p=v_e×2×sqrt2=2sqrt2 v_e .... 1
- At what height from the surface of earth the gravitation potential and the value of g are -5.4×10^7\ J kg^-2 and 6.0\ ms^-2 respectively ? Take the radius of earth as 6400 km : Answer: 1. 2600 km Key idea: V=-GM/r and g=GM/r² , so r=|V|/g . r=dfrac5.4×10⁷6.0=9.0×10⁶ m =9000 km. Height h=r-R_E=9000-6400=2600 km. 1