{"id":162781,"date":"2025-10-18T18:48:06","date_gmt":"2025-10-18T18:48:06","guid":{"rendered":"https:\/\/news.gyankatta.org\/?p=162781"},"modified":"2025-10-18T18:57:01","modified_gmt":"2025-10-18T18:57:01","slug":"class-xi-physics-gravitation-questions","status":"publish","type":"post","link":"https:\/\/news.gyankatta.org\/?p=162781","title":{"rendered":"Class XI Physics : Gravitation Questions"},"content":{"rendered":"\n<figure class=\"wp-block-image aligncenter size-large\"><img decoding=\"async\" src=\"https:\/\/manishchandra.org\/p6\/jasec2Q2gravity-min.png\" alt=\"\"\/><\/figure>\n\n\n\n<p><\/p>\n\n\n\n<div class=\"gravitation-container\">\n    <h1>Challenging Conceptual Questions: Gravitation<\/h1>\n    <p class=\"subtitle\">Class 11 Physics &#8211; Click on questions to reveal detailed answers<\/p>\n\n    <div class=\"questions-grid\">\n        <!-- Question 1 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(1)\">\n                <span class=\"question-number\">1<\/span>\n                <span class=\"question-text\">The gravitational potential energy of a body at a distance `r` is given by U(r) = -K\/r\u207f. Derive an expression for the gravitational field intensity and find the value of `n` for a realistic field.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-1\">\n                <p>The gravitational field intensity (E) is the negative gradient of potential.<\/p>\n                <p>\\[ E = -\\frac{dU}{dr} = -\\frac{d}{dr}(-Kr^{-n}) = -\\left( K n r^{-n-1} \\right) \\]<\/p>\n                <p>\\[ E = -K n r^{-(n+1)} \\]<\/p>\n                <p>For a realistic gravitational field (like Earth&#8217;s), the field must be attractive and follow the inverse-square law, meaning E \u221d -1\/r\u00b2.<\/p>\n                <p>Comparing, we get: -(n+1) = -2 \u21d2 <strong>n = 1<\/strong>.<\/p>\n                <p>Thus, for a realistic field, U(r) \u221d 1\/r, which is the standard form.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 2 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(2)\">\n                <span class=\"question-number\">2<\/span>\n                <span class=\"question-text\">A satellite is orbiting a planet of density \u03c1. Show that its orbital period T is independent of the radius of the planet.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-2\">\n                <p>For a satellite in a circular orbit of radius `r` around a planet of mass `M` and radius `R`:<\/p>\n                <p>\\[ \\frac{GMm}{r^2} = m\\omega^2 r \\]<\/p>\n                <p>Substituting \u03c9 = 2\u03c0\/T:<\/p>\n                <p>\\[ \\frac{GM}{r^2} = \\frac{4\\pi^2 r}{T^2} \\]<\/p>\n                <p>\\[ T^2 = \\frac{4\\pi^2 r^3}{GM} \\]<\/p>\n                <p>Mass of the planet M = (4\/3)\u03c0R\u00b3\u03c1. For a close orbit just above the surface, r \u2248 R:<\/p>\n                <p>\\[ T^2 = \\frac{4\\pi^2 R^3}{G \\cdot \\frac{4}{3}\\pi R^3 \\rho} = \\frac{3\\pi}{G\\rho} \\]<\/p>\n                <p>\\[ T = \\sqrt{\\frac{3\\pi}{G\\rho}} \\]<\/p>\n                <p>This shows that the period <strong>T is independent of the planet&#8217;s radius R<\/strong> and depends only on the density \u03c1.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 3 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(3)\">\n                <span class=\"question-number\">3<\/span>\n                <span class=\"question-text\">A straight, infinitely long, thin rod has mass per unit length \u03bb. Derive the expression for gravitational field intensity at a perpendicular distance `a` from the rod.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-3\">\n                <p>Consider a small element `dx` of the rod at distance `x` from the foot of the perpendicular. Mass `dm = \u03bbdx`.<\/p>\n                <p>The field due to this element is `dE = G dm \/ s\u00b2`, where `s` is distance to point P.<\/p>\n                <p>Due to symmetry, vertical components cancel. Net field is integral of horizontal components:<\/p>\n                <p>\\[ dE_{net} = \\frac{G \\lambda dx}{s^2} \\cdot \\frac{a}{s} = \\frac{G \\lambda a dx}{(a^2 + x^2)^{3\/2}} \\]<\/p>\n                <p>Integrating from x = -\u221e to \u221e:<\/p>\n                <p>\\[ E = \\int_{-\\infty}^{\\infty} \\frac{G \\lambda a dx}{(a^2 + x^2)^{3\/2}} \\]<\/p>\n                <p>Using standard integral, this gives:<\/p>\n                <p>\\[ E = G \\lambda a \\cdot \\frac{2}{a^2} = \\frac{2G\\lambda}{a} \\]<\/p>\n                <p>The field is inversely proportional to the perpendicular distance `a`.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 4 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(4)\">\n                <span class=\"question-number\">4<\/span>\n                <span class=\"question-text\">Considering Earth&#8217;s rotation, derive the expression for apparent weight at latitude \u03c6. Why is weight maximum at poles and minimum at equator?<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-4\">\n                <p>Apparent weight is the normal reaction (mg&#8217;) from Earth&#8217;s surface. True gravitational force (mg) acts towards center. A body experiences centrifugal force m\u03c9\u00b2R cos\u03c6 radially outward.<\/p>\n                <p>Resolving forces, effective weight:<\/p>\n                <p>mg&#8217; \u2248 mg &#8211; m\u03c9\u00b2R cos\u00b2\u03c6<\/p>\n                <ul>\n                    <li>At poles (\u03c6 = 90\u00b0), cos\u03c6 = 0, so mg&#8217; = mg (maximum)<\/li>\n                    <li>At equator (\u03c6 = 0\u00b0), cos\u03c6 = 1, so mg&#8217; = mg &#8211; m\u03c9\u00b2R (minimum)<\/li>\n                <\/ul>\n                <p>Weight is maximum at poles and minimum at equator due to centrifugal force being maximum at equator and zero at poles.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 5 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(5)\">\n                <span class=\"question-number\">5<\/span>\n                <span class=\"question-text\">Two stars of masses M\u2081 and M\u2082 revolve in circular orbits about their common center of mass. Prove that T = 2\u03c0 \u221a[d\u00b3 \/ G(M\u2081 + M\u2082)], where `d` is their separation.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-5\">\n                <p>Let distances from center of mass be r\u2081 and r\u2082, so d = r\u2081 + r\u2082.<\/p>\n                <p>From center of mass: M\u2081r\u2081 = M\u2082r\u2082.<\/p>\n                <p>Both stars have same angular velocity \u03c9. Gravitational force provides centripetal force:<\/p>\n                <p>For star M\u2081: (G M\u2081 M\u2082) \/ d\u00b2 = M\u2081 \u03c9\u00b2 r\u2081<\/p>\n                <p>We know r\u2081 = M\u2082 d \/ (M\u2081 + M\u2082)<\/p>\n                <p>Substituting: (G M\u2082) \/ d\u00b2 = \u03c9\u00b2 [M\u2082 d \/ (M\u2081 + M\u2082)]<\/p>\n                <p>G \/ d\u00b2 = \u03c9\u00b2 [d \/ (M\u2081 + M\u2082)]<\/p>\n                <p>\u03c9\u00b2 = G (M\u2081 + M\u2082) \/ d\u00b3<\/p>\n                <p>Since \u03c9 = 2\u03c0\/T:<\/p>\n                <p>\\[ T = 2\\pi \\sqrt{\\frac{d^3}{G(M_1 + M_2)}} \\]<\/p>\n                <p>This is Kepler&#8217;s third law for a binary system.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 6 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(6)\">\n                <span class=\"question-number\">6<\/span>\n                <span class=\"question-text\">Explain the gravitational potential energy &#8220;paradox&#8221; in an infinite universe and how defining U=0 at infinity resolves this.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-6\">\n                <p>The formula U = -GMm\/r sets U=0 at r\u2192\u221e. This works for isolated systems where gravitational influence vanishes at infinity.<\/p>\n                <p>In an infinite, homogeneous universe, gravitational potential at any point is sum of contributions from all masses. This sum doesn&#8217;t converge to a finite value &#8211; it depends on integration path, leading to undefined (infinite) potential.<\/p>\n                <p>This paradox highlights that Newtonian gravity isn&#8217;t well-suited for infinite static universes and motivated Einstein&#8217;s general relativity, where spacetime geometry describes gravity.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 7 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(7)\">\n                <span class=\"question-number\">7<\/span>\n                <span class=\"question-text\">A planet is shaped like a long, thin, uniform cylinder. Describe gravitational field intensity (E) vs. distance (r) from central axis, both inside and outside.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-7\">\n                <p><strong>Outside cylinder (r > R):<\/strong> Treated as thin rod. E \u221d 1\/r. Graph decreases like 1\/r.<\/p>\n                <p><strong>Inside cylinder (r < R):<\/strong> Using Gauss&#8217;s law for gravitation, field depends only on mass enclosed within Gaussian surface. Mass enclosed \u221d r\u00b2, so E \u221d (M_enc)\/r \u221d (r\u00b2)\/r \u221d r.<\/p>\n                <p>Graph starts from E=0 at r=0, increases linearly to maximum at r=R, then decreases as 1\/r for r>R.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 8 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(8)\">\n                <span class=\"question-number\">8<\/span>\n                <span class=\"question-text\">If Sun collapsed into a black hole of same mass, would planetary orbits change? Justify rigorously.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-8\">\n                <p><strong>No, orbits wouldn&#8217;t change significantly.<\/strong><\/p>\n                <p>Gravitational force F = GMm\/r\u00b2, where M is Sun&#8217;s mass. Since mass M remains same and planetary distances r are much larger than Schwarzschild radius of black hole, force remains identical.<\/p>\n                <p>Planets would continue in same elliptical orbits. Only change would be removal of light and radiation pressure, which have negligible effects on orbits compared to gravity.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 9 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(9)\">\n                <span class=\"question-number\">9<\/span>\n                <span class=\"question-text\">A tunnel is dug along a chord of Earth. Prove motion of particle dropped in it is simple harmonic and find its period.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-9\">\n                <p>Let chord be distance `d` from Earth&#8217;s center. At point distance `x` from midpoint, gravitational force due to mass enclosed within radius `r = \u221a(d\u00b2 + x\u00b2)`.<\/p>\n                <p>Only component along tunnel (F cos\u03b8) provides restoring force.<\/p>\n                <p>F_grav = (4\/3)\u03c0G\u03c1 m r<\/p>\n                <p>F_net = -F_grav cos\u03b8 = -[(4\/3)\u03c0G\u03c1 m r] * (x\/r) = -[(4\/3)\u03c0G\u03c1 m] x<\/p>\n                <p>Since (4\/3)\u03c0G\u03c1 is constant, F_net = -k m x \u2192 condition for SHM.<\/p>\n                <p>\u03c9\u00b2 = (4\u03c0G\u03c1\/3), T = 2\u03c0\/\u03c9 = \u221a(3\u03c0 \/ G\u03c1)<\/p>\n                <p>This period is <strong>same for all chords<\/strong> and equals period for tunnel through center.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 10 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(10)\">\n                <span class=\"question-number\">10<\/span>\n                <span class=\"question-text\">Two identical solid spheres of radius R and density \u03c1 are placed in contact. Calculate work required to disassemble them to infinity.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-10\">\n                <p>Work required = Gravitational binding energy of system = -U_total<\/p>\n                <p>Mass of each sphere M = (4\/3)\u03c0R\u00b3\u03c1<\/p>\n                <p>Distance between centers = 2R<\/p>\n                <p>Mutual P.E.: U\u2081\u2082 = -G M\u00b2 \/ (2R)<\/p>\n                <p>Self-energy of each sphere: U_self = -(3\/5)GM\u00b2\/R<\/p>\n                <p>Total initial P.E.: U_initial = 2 * [-(3\/5)GM\u00b2\/R] + [ -GM\u00b2\/(2R) ]<\/p>\n                <p>U_initial = &#8211; (6\/5 + 1\/2) GM\u00b2\/R = &#8211; (17\/10) GM\u00b2\/R<\/p>\n                <p>Work required = 0 &#8211; U_initial = <strong>(17\/10) GM\u00b2\/R<\/strong><\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 11 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(11)\">\n                <span class=\"question-number\">11<\/span>\n                <span class=\"question-text\">A space probe is projected vertically with speed \u221a(gR). What is its speed at height R above surface? (R = Earth&#8217;s radius)<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-11\">\n                <p>Using energy conservation:<\/p>\n                <p>Initial K.E. = (1\/2)m (\u221a(gR))\u00b2 = (1\/2)m gR<\/p>\n                <p>Initial P.E. = -GMm\/R = -mgR<\/p>\n                <p>Total Initial Energy = (1\/2)mgR &#8211; mgR = &#8211; (1\/2)mgR<\/p>\n                <p>At height R, distance from center = 2R<\/p>\n                <p>Final P.E. = -GMm\/(2R) = &#8211; (1\/2) mgR<\/p>\n                <p>Energy conservation: &#8211; (1\/2)mgR = (1\/2)mv\u00b2 &#8211; (1\/2)mgR<\/p>\n                <p>(1\/2)mv\u00b2 = 0 \u21d2 <strong>v = 0<\/strong><\/p>\n                <p>The probe comes to rest momentarily at height R.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 12 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(12)\">\n                <span class=\"question-number\">12<\/span>\n                <span class=\"question-text\">A planet has slight asymmetry (point mass m at distance d from center). Analyze how this affects satellite orbit, leading to precession.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-12\">\n                <p>The point mass creates non-central gravitational force. Primary force is towards planet&#8217;s center, but additional force has components perpendicular to orbit plane and within plane that aren&#8217;t purely radial.<\/p>\n                <p>This perturbing force creates torque on satellite&#8217;s angular momentum. Since \u03c4 = dL\/dt, this torque causes angular momentum vector (and orbital plane) to precess.<\/p>\n                <p>Satellite&#8217;s orbit won&#8217;t be perfect closed ellipse but will undergo <strong>nodal precession<\/strong>.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 13 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(13)\">\n                <span class=\"question-number\">13<\/span>\n                <span class=\"question-text\">Derive escape velocity from planet surface using energy conservation. Generalize for any distance r > R from center.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-13\">\n                <p>Escape velocity is minimum speed to reach infinity with zero kinetic energy.<\/p>\n                <p>Energy Conservation: (K.E. + P.E.)_surface = (K.E. + P.E.)_infinity<\/p>\n                <p>(1\/2)mv_esc\u00b2 + (-GMm\/R) = 0 + 0<\/p>\n                <p>(1\/2)mv_esc\u00b2 = GMm\/R<\/p>\n                <p><strong>v_esc = \u221a(2GM\/R)<\/strong><\/p>\n                <p>Generalizing for distance `r` from center:<\/p>\n                <p>(1\/2)mv_esc\u00b2 + (-GMm\/r) = 0<\/p>\n                <p><strong>v_esc(r) = \u221a(2GM\/r)<\/strong><\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 14 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(14)\">\n                <span class=\"question-number\">14<\/span>\n                <span class=\"question-text\">Prove gravitational field inside a uniform hollow spherical shell is zero everywhere.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-14\">\n                <p>Using shell theorem and symmetry. Consider point P inside shell. Imagine double cone with vertex at P, intersecting shell in areas dA\u2081 and dA\u2082.<\/p>\n                <p>Masses dm\u2081 = \u03c3 dA\u2081 and dm\u2082 = \u03c3 dA\u2082.<\/p>\n                <p>Fields at P: dE\u2081 = G dm\u2081 \/ r\u2081\u00b2 and dE\u2082 = G dm\u2082 \/ r\u2082\u00b2, directed towards dA\u2081 and dA\u2082.<\/p>\n                <p>From geometry, dA\u2081 \/ r\u2081\u00b2 = dA\u2082 \/ r\u2082\u00b2 (same solid angle). So dm\u2081 \/ r\u2081\u00b2 = dm\u2082 \/ r\u2082\u00b2, thus dE\u2081 = dE\u2082.<\/p>\n                <p>Since directions are opposite, they cancel exactly. Entire shell can be divided into such pairs, leading to complete cancellation.<\/p>\n                <p>Hence, <strong>net field inside hollow spherical shell is zero<\/strong>.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 15 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(15)\">\n                <span class=\"question-number\">15<\/span>\n                <span class=\"question-text\">Satellite in circular orbit at 2R fires rockets to transfer to orbit at 4R. Calculate total work done by rockets.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-15\">\n                <p>Total energy for circular orbit: E = -GMm\/(2r)<\/p>\n                <p>Initial orbit r_i = 2R: E_i = -GMm\/(4R)<\/p>\n                <p>Final orbit r_f = 4R: E_f = -GMm\/(8R)<\/p>\n                <p>Work done = Change in mechanical energy:<\/p>\n                <p>W = E_f &#8211; E_i = [-GMm\/(8R)] &#8211; [-GMm\/(4R)]<\/p>\n                <p>W = (-1\/8 + 2\/8) GMm\/R = <strong>(1\/8) GMm\/R<\/strong><\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 16 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(16)\">\n                <span class=\"question-number\">16<\/span>\n                <span class=\"question-text\">Gravitational field is E = -k\/x\u00b2. Find potential difference between x = a and x = b.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-16\">\n                <p>Relationship: E = -dV\/dx<\/p>\n                <p>dV = -E dx = -(-k\/x\u00b2) dx = k\/x\u00b2 dx<\/p>\n                <p>Integrating from x=a to x=b:<\/p>\n                <p>\u0394V = V_b &#8211; V_a = \u222b_a^b (k\/x\u00b2) dx = k [ -1\/x ]_a^b<\/p>\n                <p>\u0394V = <strong>k (1\/a &#8211; 1\/b)<\/strong><\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 17 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(17)\">\n                <span class=\"question-number\">17<\/span>\n                <span class=\"question-text\">Meteoroid at rest at large distance falls to Earth. With what speed does it strike atmosphere (100 km above surface)?<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-17\">\n                <p>Using energy conservation: Initial E = 0 (at rest at infinity)<\/p>\n                <p>Final E before atmosphere = (1\/2)mv\u00b2 &#8211; GMm\/(R_E + h)<\/p>\n                <p>0 = (1\/2)mv\u00b2 &#8211; GMm\/(R_E + h)<\/p>\n                <p>(1\/2)v\u00b2 = GM\/(R_E + h) = g R_E\u00b2 \/ (R_E + h)<\/p>\n                <p>v = \u221a[ 2g R_E\u00b2 \/ (R_E + h) ]<\/p>\n                <p>R_E = 6.4\u00d710\u2076 m, h = 1\u00d710\u2075 m, R_E + h \u2248 6.5\u00d710\u2076 m<\/p>\n                <p>v \u2248 \u221a[ 2 \u00d7 9.8 \u00d7 (6.4\u00d710\u2076)\u00b2 \/ (6.5\u00d710\u2076) ] \u2248 11,100 m\/s or <strong>11.1 km\/s<\/strong><\/p>\n                <p>Slightly less than 11.2 km\/s escape velocity because it starts from 100 km height, not surface.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 18 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(18)\">\n                <span class=\"question-number\">18<\/span>\n                <span class=\"question-text\">Why does astronaut in orbiting space station feel &#8220;weightless&#8221; even though gravity is about 90% of surface value?<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-18\">\n                <p>Weightlessness is sensation of having no apparent weight &#8211; no normal reaction force from supporting surface.<\/p>\n                <p>Astronaut and space station are both in <strong>free fall<\/strong> towards Earth with same acceleration (centripetal acceleration).<\/p>\n                <p>No relative acceleration between them, so astronaut doesn&#8217;t press against walls or floor.<\/p>\n                <p>It&#8217;s not absence of gravity, but state of continuous free fall.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 19 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(19)\">\n                <span class=\"question-number\">19<\/span>\n                <span class=\"question-text\">Geostationary satellite&#8217;s mass halves suddenly. Will it remain in same orbit, fall to Earth, or drift away? Explain.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-19\">\n                <p>Orbital velocity v = \u221a(GM\/r) and period T = 2\u03c0\u221a(r\u00b3\/GM) for circular orbit are <strong>independent of orbiting mass<\/strong>.<\/p>\n                <p>Centripetal force requirement: GMm\/r\u00b2 = mv\u00b2\/r \u2192 mass `m` cancels out.<\/p>\n                <p>If satellite halves mass without impulse (both parts continue with same velocity), remaining part will continue in <strong>exact same orbit<\/strong>.<\/p>\n                <p>It will not fall or drift away.<\/p>\n            <\/div>\n        <\/div>\n\n        <!-- Question 20 -->\n        <div class=\"question-card\">\n            <button class=\"question-btn\" onclick=\"toggleAnswer(20)\">\n                <span class=\"question-number\">20<\/span>\n                <span class=\"question-text\">A planet&#8217;s orbital period is 4 times Earth&#8217;s. What is radius of its orbit (in AU)? Assume circular orbits.<\/span>\n                <span class=\"toggle-icon\">+<\/span>\n            <\/button>\n            <div class=\"answer-content\" id=\"answer-20\">\n                <p>Using Kepler&#8217;s third law: T\u00b2 \u221d r\u00b3<\/p>\n                <p>(T_planet \/ T_earth)\u00b2 = (r_planet \/ r_earth)\u00b3<\/p>\n                <p>(4)\u00b2 = (r_planet \/ 1 AU)\u00b3<\/p>\n                <p>16 = (r_planet)\u00b3<\/p>\n                <p>r_planet = 16^(1\/3) = (2^4)^(1\/3) = 2^(4\/3) AU<\/p>\n                <p>r_planet \u2248 <strong>2.52 AU<\/strong><\/p>\n            <\/div>\n        <\/div>\n    <\/div>\n<\/div>\n\n<style>\n.gravitation-container {\n    max-width: 900px;\n    margin: 0 auto;\n    padding: 20px;\n    font-family: 'Segoe UI', Tahoma, Geneva, Verdana, sans-serif;\n    background: linear-gradient(135deg, #667eea 0%, #764ba2 100%);\n    border-radius: 15px;\n    box-shadow: 0 10px 30px rgba(0,0,0,0.2);\n}\n\n.gravitation-container h1 {\n    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Derive an expression for the gravitational field intensity and find the value of `n` for a realistic field. + The gravitational field [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":162783,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"fifu_image_url":"https:\/\/manishchandra.org\/p6\/jasec2Q2gravity-min.png","fifu_image_alt":"","footnotes":""},"categories":[3,6],"tags":[],"class_list":["post-162781","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-education","category-exam","cat-3-id","cat-6-id","has_thumb"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v27.3 - https:\/\/yoast.com\/product\/yoast-seo-wordpress\/ -->\n<title>Class XI Physics : Gravitation Questions - Gyankatta<\/title>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/news.gyankatta.org\/?p=162781\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Class XI Physics : Gravitation Questions - Gyankatta\" \/>\n<meta property=\"og:description\" content=\"Challenging Conceptual Questions: Gravitation Class 11 Physics &#8211; Click on questions to reveal detailed answers 1 The gravitational potential energy of a body at a distance `r` is given by U(r) = -K\/r\u207f. 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