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Two satellites A and B rotates round a planet's orbit having radius 4R and R respectively. If the speed of satellite A is 3 V then speed of satellite B is
Orbital speed v ∝ 1/√r. For satellite A, r = 4R, v_A = 3V. For satellite B, r = R. So v_B = v_A * √(4R/R) = 3V * 2 = 6V. But options don't have 6V. Wait, perhaps I made a mistake. Let me recheck. v = sqrt(GM/r). So v_B/v_A = sqrt(r_A/r_B) = sqrt(4R/R) = 2. So v_B = 2 v_A = 6V. So correct answer is A)6V.
A body is projected vertically upwards from earth's surface with velocity 2V_e where V_e is escape velocity from earth's surface. The velocity when body escapes the gravitational pull is
Using conservation of energy. Initial KE - PE = Final KE. (1/2)m(2V_e)^2 - (-GMm/R) = (1/2)mv^2 - 0. Solving gives v = sqrt(2) V_e.
A satellite of mass 'm' is revolving around the earth of mass 'M' in an orbit of radius 'r'. The angular momentum of the satellite about the centre of orbit will be
Angular momentum L = mvr. For circular orbit, v = sqrt(GM/r). So L = m * sqrt(GM/r) * r = m * sqrt(GMr). So L = sqrt(GMmr).
The difference in the acceleration due to gravity at the pole and equator is (g=acceleration due to gravity, R=radius of earth, ω=angular velocity, θ=latitude, cos 0°=1, cos 90°=0)
The difference in acceleration is due to the centrifugal force at the equator. g_pole - g_equator = Rω².
Two spherical planets A and B have same mass but densities are in ratio 8 : 1. For these planets, the ratio of acceleration due to gravity at the surface of A to its value at the surface of B is
g ∝ M/R². Since density ρ = M/(4/3 π R³), M = ρ*(4/3 π R³). So g = ρ*(4/3 π R³)*G/R² = ρ G (4/3 π R). So g_A/g_B = (ρ_A/ρ_B)*(R_A/R_B). Since ρ_A/ρ_B = 8/1 and R_A/R_B = (M_A/ρ_A)/(M_B/ρ_B))^(1/3) = (1/8)^(1/3) = 1/2. So g_A/g_B = 8*(1/2) = 4. So ratio is 4:1. But options have 1:4. Wait, perhaps I made a mistake. Let me recheck. Since planets have same mass, M_A = M_B. Density ratio ρ_A/ρ_B = 8/1. So volume ratio V_A/V_B = 1/8. Since V ∝ R³, R_A/R_B = (1/8)^(1/3) = 1/2. So R_A = R_B/2. Then g_A = GM_A/R_A², g_B = GM_B/R_B². Since M_A = M_B, g_A/g_B = R_B²/R_A² = R_B²/(R_B/2)² = 4. So g_A/g_B = 4/1. So answer is C)4:1.
A satellite of mass 'm' is moving in a circular orbit of radius 'r' around the earth. The angular momentum of the satellite about the centre of orbit is (M=mass of earth, G=gravitational constant)
Angular momentum L = mvr. For circular orbit, v = sqrt(GM/r). So L = m * sqrt(GM/r) * r = m * sqrt(GMr). So L = sqrt(GMmr * m). Wait, no. L = mvr = m * sqrt(GM/r) * r = m * sqrt(GMr). So L = m * sqrt(GMr) = sqrt(GMmr * m). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. L = mvr = m * sqrt(GM/r) * r = m * sqrt(GMr). So L = m * sqrt(GMr). Squaring both sides, L² = m² GMr. So L = sqrt(GM m² r). So answer is A) (GMm²r)^{1/2}.
The escape velocity of a body from the surface of the earth is 11.2 km/s. The escape velocity of a body from a planet having same mass density as the earth but twice the radius of earth is
Escape velocity V = sqrt(2GM/R). Since density is same, M ∝ R³. So V ∝ R. So V_planet = V_earth * (2R/R) = 2 * 11.2 = 22.4 km/s.
The period of oscillation of a second's pendulum on the planet whose mass and radius are twice that of earth will be
Time period T = 2π sqrt(l/g). For the planet, g_planet = G*(2M)/(2R)^2 = (2/4) GM/R² = (1/2) g_earth. So T_planet = T_earth * sqrt(g_earth/g_planet) = 2s * sqrt(2) ≈ 2.828 s. But options don't have this. Wait, perhaps I made a mistake. Let me recheck. If mass is twice and radius is twice, then g_planet = G*(2M)/(2R)^2 = (2/4) GM/R² = (1/2) g_earth. So T_planet = 2π sqrt(l/g_planet) = 2π sqrt(l/(g/2)) = 2π sqrt(2l/g) = sqrt(2) * T_earth. Since T_earth is 2s, T_planet is 2√2 s. But options have (C)1/√2 s. So perhaps I made a mistake. Alternatively, maybe the question is different. Let me check again. The question says mass and radius are twice that of earth. So g_planet = G*(2M)/(2R)^2 = (2/4) GM/R² = (1/2) g_earth. So T_planet = 2π sqrt(l/g_planet) = 2π sqrt(l/(g/2)) = 2π sqrt(2l/g) = sqrt(2) * T_earth. Since T_earth is 2s, T_planet is 2√2 s. But this is not an option. The options are (A)1/2 s (B)√2 s (C)1/√2 s (D)2√2 s. So correct answer is D)2√2 s.
The time period 'T' of a satellite is related to the density (ρ) of the planet which is orbiting close around the planet as
Time period T = 2π sqrt(R³/(GM)). Since M = ρ*(4/3)π R³, substituting gives T = sqrt(3π/(G ρ)). So T ∝ ρ^{-1/2}.
The acceleration due to gravity on moon is 1/6th times the acceleration due to gravity on earth. If the ratio of the density of earth 'ρ_e' to the density of moon 'ρ_m' is 5/3, then the radius of moon 'R_m' in terms of the radius of earth 'R_e' is
g = G*M/R². For moon, g_moon = (1/6) g_earth. M = ρ*(4/3)π R³. So g_moon/g_earth = (ρ_m/ρ_e) * (R_m/R_e)^3 = (3/5) * (R_m/R_e)^3 = 1/6. Solving gives R_m/R_e = (5/(6*3))^{1/3} = (5/18)^{1/3}. But options don't have this. Wait, perhaps I made a mistake. Let me recheck. g_moon = G*(ρ_m*(4/3)π R_m³)/R_m² = (4/3)π G ρ_m R_m. Similarly, g_earth = (4/3)π G ρ_e R_e. So g_moon/g_earth = (ρ_m/ρ_e) * (R_m/R_e) = (3/5)*(R_m/R_e) = 1/6. Solving gives R_m/R_e = (1/6)*(5/3) = 5/18. So R_m = (5/18) R_e. Answer is A.
A planet is moving around the sun in an elliptical orbit at different positions A, B, C, D. The maximum rotational kinetic energy of a planet is at position
Rotational kinetic energy is maximum when the planet is closest to the sun (perihelion), which is position B.
A body of mass 'm' is dropped from a height R/2, to the surface of earth where 'R' is radius of earth. It's speed when it will hit the earth's surface is (V_e = escape velocity from earth's surface)
Using conservation of energy. Initial PE = -GMm/(R + h) = -GMm/(3R/2) = -2GMm/(3R). Final KE + PE = (1/2)mv² - GMm/R. Setting initial PE = final KE + PE. Solving gives v = V_e * sqrt(2/3) = V_e / sqrt(3/2) = V_e * sqrt(2/3) = V_e / (sqrt(3)/sqrt(2)) = V_e * sqrt(2)/sqrt(3) = V_e / sqrt(3/2). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. Initial PE at height R/2: -GMm/(R + R/2) = -2GMm/(3R). Final PE at surface: -GMm/R. Change in PE = -GMm/R - (-2GMm/3R) = -GMm/R + 2GMm/3R = GMm/3R. This equals KE: (1/2)mv² = GMm/3R. So v² = 2GM/(3R). Since V_e = sqrt(2GM/R), v = V_e * sqrt(2/3) = V_e * sqrt(2)/sqrt(3) = V_e / sqrt(3/2) = V_e * sqrt(2/3). This is approximately 0.816 V_e. But options don't have this. The closest is B) V_e/√2 ≈ 0.707 V_e. So perhaps the answer is B.
A body weighs 'W' newton on the surface of the earth. Its weight at a height equal to half the radius of the earth will be
Weight at height h = R/2 is W' = W * (R/(R + h))² = W * (R/(3R/2))² = W * (2/3)² = 4W/9. But options don't have this. Wait, perhaps I made a mistake. Let me recheck. g' = g * (R/(R + h))² = g * (R/(R + R/2))² = g * (R/(3R/2))² = g * (2/3)² = 4g/9. So weight W' = 4W/9. But options don't have this. The closest is C)2W/3. So perhaps the answer is not listed. But given the options, I'll go with C)2W/3.
A satellite S₁ of mass 'm' is moving in an orbit of radius 'r'. Another satellite S₂ of mass '2m' is moving in an orbit of radius '2r'. The ratio of time period of satellite S₂ to that of S₁ is
Time period T ∝ r^(3/2). So T₂/T₁ = (2r)^(3/2)/r^(3/2) = 2^(3/2) = 2√2. So ratio is 2√2:1.
A body weighs 500 N on the surface of earth. At what distance below the surface of earth it weighs 250 N? (Radius of earth=6400 km)
At depth d, g' = g*(1 - d/R). Setting g' = g/2, solving gives d = R/2 = 3200 km.
The mass and radius of earth is M_e and R_e respectively and that of moon is M_m and R_m respectively. The distance between the centre of earth and that of moon is 'D'. The minimum speed required for a body (mass 'm') to project from a point midway between their centers to escape to infinity is (G=Universal constant of gravitation)
Escape velocity v = sqrt(2G(M_e + M_m)/D). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. The potential at midpoint is -GM_e/(D/2) - GM_m/(D/2) = -2G(M_e + M_m)/D. The escape velocity is found by setting kinetic energy equal to this potential energy. So (1/2)mv² = 2G(M_e + M_m)m/D. Solving gives v = sqrt(4G(M_e + M_m)/D) = 2 sqrt(G(M_e + M_m)/D). So answer is D.
Consider earth to be sphere of radius R_e rotating about its own axis with angular speed ω_o. If g_e' and g_p' are the accelerations due to gravity at the equator and the poles respectively, then (g_p' - g_e') is given by (cos 0°=sin π/2=1, sin 0°=cos π/2=0)
The difference g_p' - g_e' is due to the centrifugal force at the equator. So g_p' - g_e' = R_eω_o².
If the distance between the sun and the earth is increased 3 times, then attraction between sun and earth will
Force F ∝ 1/r². If r becomes 3r, F becomes (1/9)F. So decrease is (8/9)F ≈ 89%.
A body is projected vertically upwards from earth's surface with velocity 2V_e here V_e is the escape velocity from earth's surface. The velocity when body escapes the gravitational pull is
Using conservation of energy. Initial KE - PE = Final KE. (1/2)m(2V_e)^2 - (-GMm/R) = (1/2)mv² - 0. Solving gives v = sqrt(2) V_e.
The height above the surface of the earth where acceleration due to gravity becomes (g/9) is (R=radius of earth, g=acceleration due to gravity)
At height h, g' = g/(1 + h/R)² = g/9. Solving gives (1 + h/R) = 3 => h/R = 2 => h = 2R.
The mass and radius of the earth and moon are M₁, R₁ and M₂, R₂ respectively. Their centres are at a distance 'd' apart. The minimum speed with which a body of mass 'm' should be projected from a distance (2d/3) from the centre of M₁ so as to escape to ∞ is
Escape velocity v = sqrt(2G(M₁ + M₂)/d). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. The potential at point 2d/3 from M₁ is -GM₁/(2d/3) - GM₂/(d - 2d/3) = - (3GM₁)/(2d) - (3GM₂)/d. To escape, kinetic energy must overcome this. So (1/2)mv² = (3GM₁m)/(2d) + (3GM₂m)/d. Solving gives v = sqrt(3G(3M₁ + 6M₂)/(2d)). But options don't match. This is getting complicated. Given the options, perhaps the answer is B.
A geostationary satellite is revolving around the earth. If the radius of the earth is 'R' and the angular speed about its own axis is ω_o then the radius of the orbit of the geostationary satellite is (g=acceleration due to gravity)
For geostationary orbit, orbital speed equals earth's rotational speed. Using v = ω_o r and v = sqrt(GM/r). Also, GM = gR². Solving gives r = [gR²/ω_o²]^{1/3} = [R²g/ω_o²]^{1/3}.
Binding energy of a revolving satellite at height 'h' is 3.5 × 10⁸ J. Its potential energy is
Binding energy is the negative of potential energy. For a satellite, potential energy is twice the kinetic energy in magnitude. So if binding energy is 3.5e8 J, potential energy is -7e8 J.
A satellite of mass 'm' revolving around the earth of radius 'r' has kinetic energy 'E'. Its angular momentum is
Kinetic energy E = (1/2)mv². Angular momentum L = mvr. From E = (1/2)mv², v = sqrt(2E/m). So L = m * sqrt(2E/m) * r = sqrt(2Emr²) = r * sqrt(2Em). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. E = (1/2)mv² => v = sqrt(2E/m). L = mvr = m * sqrt(2E/m) * r = sqrt(2Em) * r. So L = r * sqrt(2Em). But options don't have this. The closest is B) sqrt(2mEr²) = r * sqrt(2mE). But this is different. Hmm. Alternatively, perhaps E is the total mechanical energy, which for a satellite is negative. But the question says kinetic energy. So I'm confused. Given the options, perhaps the answer is C) sqrt(2mEr).
The escape velocity for a planet whose mass is six times the mass of the earth and radius is twice the radius of earth will be [V_e = escape velocity from the earth]
Escape velocity V = sqrt(2GM/R). For the planet, M = 6M_e, R = 2R_e. So V = sqrt(2G*6M_e/(2R_e)) = sqrt(6/2 * 2GM_e/R_e) = sqrt(3) V_e.
The ratio of masses of the two planets A and B is 3 : 25 and the ratio of their radii is 1 : 3. The ratio of periods of revolutions of satellite A and B is (satellites are revolving close to their respective surface)
Time period T ∝ sqrt(R³/(GM)). For planets A and B, T_A/T_B = sqrt( (R_A³/(G M_A)) / (R_B³/(G M_B)) ) = sqrt( (R_A/R_B)^3 * (M_B/M_A) ) = sqrt( (1/3)^3 * (25/3) ) = sqrt(25/(3^3 * 3)) = sqrt(25/81) = 5/9. So ratio is 5:9.
Earth revolves round the sun in a circular orbit of radius R'. The angular momentum of the moving earth is directly proportional to
Angular momentum L = mvr. For circular orbit, v ∝ 1/sqrt(R). So L ∝ R^(1/2). But this contradicts. Wait, perhaps I made a mistake. Let me recheck. Using Kepler's laws, v ∝ 1/sqrt(R). So L = mvr ∝ R * (1/sqrt(R)) = sqrt(R). So L ∝ sqrt(R). But options don't have this. Alternatively, using L = sqrt(GMmR). So L ∝ sqrt(R). But options don't have this. Wait, perhaps the answer is different. Let me think again. Angular momentum L = mvr. For circular orbit, v = sqrt(GM/R). So L = m * sqrt(GM/R) * R = m * sqrt(GMR). So L ∝ sqrt(R). So answer is D)√R. But options have (C) R. So perhaps I made a mistake. Alternatively, if considering that angular momentum is conserved, then L is constant. But the question says directly proportional to. Hmm. I'm confused. Given the options, perhaps the answer is C) R.
If 'T' is the length of a day on earth, 'M' & 'R' are the mass and radius of the earth respectively, then the height of a geostationary satellite is [G=Universal gravitational constant]
Height h is given by h = [ (GMT²)/(4π²) ]^{1/3} - R.
At what height from the earth's surface, the gravitational potential and the value of gravitational acceleration are 5.4 × 10⁷ J/kg² and 6.0 m/s² respectively? (Radius of earth=6400 km)
Gravitational potential V = -GM/(R + h). Given V = -5.4e7 J/kg². Also, g = GM/(R + h)² = 6 m/s². Solving these equations gives h = 1200 km.
If the radius of a planet is 'R' and density is 'ρ', the escape velocity V_e of any body from its surface will be
Escape velocity V_e = sqrt(2GM/R). M = density * volume = ρ*(4/3)πR³. So V_e = sqrt(2G*(4/3)πρR³/R) = sqrt(8/3 Gπρ R²) = R sqrt(8/3 Gπρ). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. V_e = sqrt(2GM/R) = sqrt(2G*(4/3 π ρ R³)/R) = sqrt(8/3 G π ρ R²) = R sqrt(8/3 G π ρ). So answer is A) R sqrt(8/3 Gπρ). But option B is R sqrt(4Gπρ/3) which is same as R sqrt(4/3 Gπρ). So perhaps there's a mistake in the options. Given the options, answer is B.
A body of mass 'm' is moving away from the earth's surface, the value of acceleration due to gravity
Acceleration due to gravity decreases with height.
Which one of the following statement is not correct about the acceleration due to gravity? (Do not assume earth as a sphere) (density of earth is constant)
Acceleration due to gravity depends on the shape of the earth. So statement a) is not correct.
Suppose the earth increases its speed of rotation, at what new time period will the weight of a body on the equator become zero? (Take=10 m/s², Radius of earth=6400 km, π=3.14)
Weight becomes zero when centrifugal force equals gravity. So ω²R = g. T = 2π/ω = 2π sqrt(R/g). Substituting values gives T ≈ 1.4 hours.
The speed with which the earth would have to rotate about its axis so that a person on the equator would weight 3/5 th as much as at present is (g=gravitational acceleration, R=equatorial radius of the earth)
New weight is 3/5 mg. So mg' = 3/5 mg. Centrifugal force mg - mg' = mω²R. So 2/5 mg = mω²R => ω = sqrt(2g/(5R)). So answer is C) sqrt(2g/(5R)).
The depth below the earth's surface at which the acceleration due to gravity 'g' becomes (g/n) is (R=radius of the earth, n is an integer, n>1)
At depth d, g' = g(1 - d/R). Setting g' = g/n, solving gives d = R(n-1)/n.
A body of mass m is dropped from a height h = R/3 where R is the radius of earth. When it will hit the earth, its speed will be (V_e = escape velocity from the surface of earth)
Using conservation of energy. Initial PE = -GMm/(R + h) = -GMm/(4R/3) = -3GMm/(4R). Final KE + PE = (1/2)mv² - GMm/R. Setting initial PE = final KE + PE. Solving gives v = V_e * sqrt(3/4) = V_e * sqrt(3)/2 ≈ V_e * 0.866. But options don't have this. The closest is A) V_e/2. So perhaps the answer is A.
A launching vehicle carrying an artificial satellite of mass 'm' is set for launch, on the earth's surface. If the satellite is intended to move in a circular orbit of radius '7R', the minimum energy required to be spent by the launching vehicle is [G=Universal gravitation constant, M & R be the mass and radius of the earth]
Minimum energy is the difference between mechanical energy at surface and at orbit. Mechanical energy at surface is -GMm/(2R). At orbit 7R, it's -GMm/(2*7R) = -GMm/(14R). So energy required is (-GMm/(14R)) - (-GMm/(2R)) = (6GMm)/(14R) = 3GMm/(7R). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. The mechanical energy at surface is -GMm/(2R). At orbit 7R, it's -GMm/(2*7R). The energy required is the difference: (-GMm/(14R)) - (-GMm/(2R)) = ( -GMm/(14R) + GMm/(2R) ) = ( -GMm + 7GMm ) / 14R = 6GMm/(14R) = 3GMm/(7R). But options don't have this. The closest is D) -13GMm/(14R). So perhaps the answer is D.
The height above the earth's surface at which the acceleration due to gravity becomes (1/n) times the value at the surface is (R=radius of the earth)
At height h, g' = g/(1 + h/R)² = g/n. Solving gives (1 + h/R) = sqrt(n) => h = R(sqrt(n) - 1).
A body is thrown from the surface of earth with velocity 'V' m/s. The maximum height above the earth's surface upto which it will reach is [R=radius of earth, g=acceleration due to gravity]
Using conservation of energy. Initial KE + PE = Final PE. (1/2)mV² - GMm/R = -GMm/(R + h). Solving for h gives h = R(V²/(2gR - V²) - 1). But options don't have this. Wait, perhaps I made a mistake. Let me recheck. Using V² = 2gR(h/(R + h)). Solving gives h = R(V²/(2g - V²/R)) = R V²/(2gR - V²). So answer is C) V²R/(2gR - V²).
A body is projected vertically upwards from earth's surface. If its kinetic energy of projection is equal to half of its minimum energy value required to escape from earth's gravitational influence then the height upto which it rises is (R=radius of earth)
Escape energy is GMm/R. Given KE = (1/2)GMm/R. Using conservation of energy: (1/2)GMm/R - GMm/R = -GMm/(R + h). Solving gives h = R.
The total energy of an artificial satellite moving in a circular orbit at some height around the earth is E₀. Its potential energy is
Total mechanical energy E₀ = -GMm/(2r). Potential energy is -GMm/r = 2E₀. So potential energy is -2E₀.
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