PHYSICS-15-10- 11th (PQRS) SOLUTION

https://docs.google.com/document/d/1P_3cloNnI8-1h-af5lH1wrDmmTcCrvkC/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=t variation with altitude and depth. Kepler’s laws of planetary motion. Gravitational potential energy; gravitational potential. Escape velocity. Orbital velocity of a satellite. Geo-stationary satellites. UNIVERSAL LAW OF GRAVITATION Newton’s Law of Gravitation Gravitational force between two points masses or spherically symmetric m1 and m2 separated by distance r is F = Gm1m2 . r 2 G = Universal gravitational constant = 6.67 × 10–11 Nm2kg–2, [G] = [M–1L3T–2] ACCELERATION DUE TO GRAVITY Acceleration produced in a body due to earth’s gravitational pull is called acceleration due to gravity. As gravitational force on a body of mass m placed at the surface of earth is F = GMm R 2 THIS CHAPTER COVERS : Universal law of gravitation Acceleration due to gravity Gravitational potential Gravitational Potential Energy Escape Velocity Kepler’s laws Satellites Binary Star System g = F m = GM R 2 Relation between g and G : G × 4 πR3ρ g = GM = 3 R 2 R 2 = 4 πGRρ 3 ρ → density of earth = 5.5 × 103 kg/m3 TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Variation in the value of g : At height ‘h’ g′ = g and g′ = ⎡ − 2h ⎤ if h << R ⎡ h ⎤ 2 g ⎢1 ⎣ ⎥ e Re ⎦ ⎢1 + ⎥ ⎣ e ⎦ If h = Re , g′ = g 4 At depth ‘x’ g ′ = ⎧ x g ⎨1 − R ⎫ ⎬ , at the centre of earth g′ = 0, weight = 0 ⎩ e ⎭ Due to Rotation of Earth : g′ = g – Reω2cos2λ λ → angle of latitude gpole → maximum and unaffected from rotation. gequator → minimum and becomes maximum if earth stops rotation. GRAVITATIONAL FIELD INTENSITY AND POTENTIAL (V) : Gravitational field intensity at a point is the force experienced by a unit mass placed at that point I = −GM rˆ r 2 Gravitational potential at a point is the amount of work done in bringing a unit mass from infinity to that point in the gravitational field. V = − W m ⇒ V = −GM r (units J/kg) Variation of Intensity and Potential For a spherical shell of mass M and radius R Case-I : r < R (internal point) Ii = 0, Vi = GM R Case-II : r < R (on the surface I = − GM , V = −GM s R 2 s R Case-III : r > R (outside the shell) Io = GM , V = −GM r 2 o r I V O r O r GM R For Uniform solid Sphere Case-I : r < R (internal point) I = GMr , V = −GM (3R 2 − r 2 ) At centre V = − 3GM = 3 V i R 3 i 2R 3 c 2R 2 s Case-II : r = R (on the surface) I = − GM , V s R 2 s = −GM R Case-III : r > R (outside the surface) I Imax I = GM , V o r 2 o = −GM r O ` Gravitational intensity and potential on the axis of uniform ring of mass M radius R at distance x from centre. I = , V = At centre I = 0 and I is maximum at x = ± R ; Imax = 2GM The point P at which gravitational field is zero between two heavenly bodies. r r 1 2 GRAVITATIONAL POTENTIAL ENERGY Gravitational potential energy of a body at a point w.r.t earth is defined as negative of work done by gravity in bringing that body from infinity to the point of consideration. At earth surface U = − GMem , At height h, U = −GMem Re Re + h If a body of mass m is raised to a height h from the surface of earth, then gain of its potential energy is ΔU = mgRh . R + h Energy required to escape = Escape energy = + ESCAPE VELOCITY GMem Re Escape velocity on the surface of planet is the minimum velocity required to escape the body from the gravitational field of planet. To escape a body its mechanical energy should be greater than or equal to zero. 1 2 ⎛ − GMem ⎞ mve + ⎜ ⎝ Re ⎟ = 0 ⎠ KEPLER’S LAWS All planets revolve around the Sun in elliptical orbit having the Sun at one focus. If e = eccentricity of ellipse then distance of the planet from the Sun at perigee. rp = (1 – e)a and distance of the planet from the Sun at apogee ra = (1 + e)a (a = semi major axis) Ratio of orbital speeds at apogee and perigee are va = rp = 1− e v p ra 1+ e Ratio of angular velocities at apogee and perigee are w ⎛ r ⎞2 ⎛ 1− e ⎞2 a = ⎜ p ⎟ = ⎜ ⎟ w p ⎝ ra ⎠ ⎝ 1+ e ⎠ 2a A planet sweeps out equal area in equal time interval i.e. Areal speed of the planet is constant dA = 1 vr = L = constant (L = mvr = angular momentum of planet about the Sun) dt 2 2m Angular momentum about the Sun ‘L’ of all planets and satellites is constant Square of time period is proportional to cube of semi-major axis of the elliptical orbit of the planet. i.e., T 2 ∝ a3 SATELLITES Important results regarding satellite motion in circular orbit. Orbital Velocity (v0) : Gravitational attraction of planet gives necessary centripetal force. GMm r 2 ⇒ v0 = mv 2 = 0 r v 0 = = (h = height above the surface of earth) v 0 ≈ ≈ Since ve = ⇒ ve = 2v0 Time Period : The period of revolution of a satellite is T = 2πr v0 = 2πr = 2π For a satellite orbiting close to the earth’s surface (r ~ Re), the time period is minimum and is given by Tmin = 2π = 2π For earth Re = 6400 km, g = 9.8 m/s2 Tmin = 84.6 min. Thus, for any satellite orbiting around the earth, its time period must be more than 2π or 84.6 minutes. Potential Energy (U), Kinetic Energy (K) and Total Energy (E) of satellite U = − GMm r K = GMm 2r E = − GMm 2r [K = –E & U = 2E] Bound and Unbound Trajectories Imagine a very tall building on the surface of the earth from where a projectile is fired with a velocity v parallel to the surface of the earth. The trajectory of the projectile depends on its velocity. v ve Parabola ve 0 v0 < v < ve Circle (ve = escape velocity) (v0 = orbital velocity, ve = 2v 0 ) Ellipse Velocity Trajectory v < v0 Projectile may not orbit the earth in an ellipical path, or it may falls back on the earth’s surface. v = v0 Projectile orbits the earth in a circular path. v0 < v < ve Projectile orbits in an elliptical path. v = ve Projectile does not orbit. It escapes the gravitational field of earth in a parabolic path. v > ve Projectile does not orbit. It escapes the gravitational field of earth in a hyperbolic path. Geostationary Satellites The plane of the orbit lies in equatorial plane of earth. Height from the earth surface is 36000 km. This orbit is called parking orbit. Orbital speed is nearly 3 km/s. Time period is equal to that of earth rotation i.e., 24 hours. Polar Satellite Used for remote sensing, their orbit contains axis of rotation of earth. They can cover entire earth surface for viewing. BINARY STAR SYSTEM Two stars of mass M1 and M2 form a stable system when they move in circular orbit about their centre of mass under their mutual gravitational attraction. V2 M2 Applications If a body is projected with velocity greater than escape velocity, the intersteller speed is given by vI.S = If a body is projected upwards with velocity v = Kve, where K < 1, maximum height attained from centre R of earth ‘r’ = 1 − K 2 . RK 2 v 2R From surface of the earth h = r − R = 1 − K 2 = v 2 − v 2 If a body is dropped from rest form height h comparable to radius of earth, then maximum velocity with which it strikes the ground v = v e Time period of satellite is independent of mass of the satellite. Atmosphere on a planet is possible only if vrms < ve. Where vrms = rms speed of gas molecules. If angular velocity of earth is made 17 times, object placed at equator fly off. As we move from pole to equator, gravity decreases by 0.35%. If orbital velocity of any satellite very close to Earth surface increases by 41.4%, then the satellite will escape to infinity. If gravitation force ∝ rn, time period of a satellite ∝ r(1 – n)/2. ❑ ❑ ❑ I ∝r I ∝ 1 r 2 r = R variation with altitude and depth. Kepler’s laws of planetary motion. Gravitational potential energy; gravitational potential. Escape velocity. Orbital velocity of a satellite. Geo-stationary satellites. UNIVERSAL LAW OF GRAVITATION Newton’s Law of Gravitation Gravitational force between two points masses or spherically symmetric m1 and m2 separated by distance r is F = Gm1m2 . r 2 G = Universal gravitational constant = 6.67 × 10–11 Nm2kg–2, [G] = [M–1L3T–2] ACCELERATION DUE TO GRAVITY Acceleration produced in a body due to earth’s gravitational pull is called acceleration due to gravity. As gravitational force on a body of mass m placed at the surface of earth is F = GMm R 2 THIS CHAPTER COVERS : Universal law of gravitation Acceleration due to gravity Gravitational potential Gravitational Potential Energy Escape Velocity Kepler’s laws Satellites Binary Star System g = F m = GM R 2 Relation between g and G : G × 4 πR3ρ g = GM = 3 R 2 R 2 = 4 πGRρ 3 ρ → density of earth = 5.5 × 103 kg/m3 TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616 Variation in the value of g : At height ‘h’ g′ = g and g′ = ⎡ − 2h ⎤ if h << R ⎡ h ⎤ 2 g ⎢1 ⎣ ⎥ e Re ⎦ ⎢1 + ⎥ ⎣ e ⎦ If h = Re , g′ = g 4 At depth ‘x’ g ′ = ⎧ x g ⎨1 − R ⎫ ⎬ , at the centre of earth g′ = 0, weight = 0 ⎩ e ⎭ Due to Rotation of Earth : g′ = g – Reω2cos2λ λ → angle of latitude gpole → maximum and unaffected from rotation. gequator → minimum and becomes maximum if earth stops rotation. GRAVITATIONAL FIELD INTENSITY AND POTENTIAL (V) : Gravitational field intensity at a point is the force experienced by a unit mass placed at that point I = −GM rˆ r 2 Gravitational potential at a point is the amount of work done in bringing a unit mass from infinity to that point in the gravitational field. V = − W m ⇒ V = −GM r (units J/kg) Variation of Intensity and Potential For a spherical shell of mass M and radius R Case-I : r < R (internal point) Ii = 0, Vi = GM R Case-II : r < R (on the surface I = − GM , V = −GM s R 2 s R Case-III : r > R (outside the shell) Io = GM , V = −GM r 2 o r I V O r O r GM R For Uniform solid Sphere Case-I : r < R (internal point) I = GMr , V = −GM (3R 2 − r 2 ) At centre V = − 3GM = 3 V i R 3 i 2R 3 c 2R 2 s Case-II : r = R (on the surface) I = − GM , V s R 2 s = −GM R Case-III : r > R (outside the surface) I Imax I = GM , V o r 2 o = −GM r O ` Gravitational intensity and potential on the axis of uniform ring of mass M radius R at distance x from centre. I = , V = At centre I = 0 and I is maximum at x = ± R ; Imax = 2GM The point P at which gravitational field is zero between two heavenly bodies. r r 1 2 GRAVITATIONAL POTENTIAL ENERGY Gravitational potential energy of a body at a point w.r.t earth is defined as negative of work done by gravity in bringing that body from infinity to the point of consideration. At earth surface U = − GMem , At height h, U = −GMem Re Re + h If a body of mass m is raised to a height h from the surface of earth, then gain of its potential energy is ΔU = mgRh . R + h Energy required to escape = Escape energy = + ESCAPE VELOCITY GMem Re Escape velocity on the surface of planet is the minimum velocity required to escape the body from the gravitational field of planet. To escape a body its mechanical energy should be greater than or equal to zero. 1 2 ⎛ − GMem ⎞ mve + ⎜ ⎝ Re ⎟ = 0 ⎠ KEPLER’S LAWS All planets revolve around the Sun in elliptical orbit having the Sun at one focus. If e = eccentricity of ellipse then distance of the planet from the Sun at perigee. rp = (1 – e)a and distance of the planet from the Sun at apogee ra = (1 + e)a (a = semi major axis) Ratio of orbital speeds at apogee and perigee are va = rp = 1− e v p ra 1+ e Ratio of angular velocities at apogee and perigee are w ⎛ r ⎞2 ⎛ 1− e ⎞2 a = ⎜ p ⎟ = ⎜ ⎟ w p ⎝ ra ⎠ ⎝ 1+ e ⎠ 2a A planet sweeps out equal area in equal time interval i.e. Areal speed of the planet is constant dA = 1 vr = L = constant (L = mvr = angular momentum of planet about the Sun) dt 2 2m Angular momentum about the Sun ‘L’ of all planets and satellites is constant Square of time period is proportional to cube of semi-major axis of the elliptical orbit of the planet. i.e., T 2 ∝ a3 SATELLITES Important results regarding satellite motion in circular orbit. Orbital Velocity (v0) : Gravitational attraction of planet gives necessary centripetal force. GMm r 2 ⇒ v0 = mv 2 = 0 r v 0 = = (h = height above the surface of earth) v 0 ≈ ≈ Since ve = ⇒ ve = 2v0 Time Period : The period of revolution of a satellite is T = 2πr v0 = 2πr = 2π For a satellite orbiting close to the earth’s surface (r ~ Re), the time period is minimum and is given by Tmin = 2π = 2π For earth Re = 6400 km, g = 9.8 m/s2 Tmin = 84.6 min. Thus, for any satellite orbiting around the earth, its time period must be more than 2π or 84.6 minutes. Potential Energy (U), Kinetic Energy (K) and Total Energy (E) of satellite U = − GMm r K = GMm 2r E = − GMm 2r [K = –E & U = 2E] Bound and Unbound Trajectories Imagine a very tall building on the surface of the earth from where a projectile is fired with a velocity v parallel to the surface of the earth. The trajectory of the projectile depends on its velocity. v ve Parabola ve 0 v0 < v < ve Circle (ve = escape velocity) (v0 = orbital velocity, ve = 2v 0 ) Ellipse Velocity Trajectory v < v0 Projectile may not orbit the earth in an ellipical path, or it may falls back on the earth’s surface. v = v0 Projectile orbits the earth in a circular path. v0 < v < ve Projectile orbits in an elliptical path. v = ve Projectile does not orbit. It escapes the gravitational field of earth in a parabolic path. v > ve Projectile does not orbit. It escapes the gravitational field of earth in a hyperbolic path. Geostationary Satellites The plane of the orbit lies in equatorial plane of earth. Height from the earth surface is 36000 km. This orbit is called parking orbit. Orbital speed is nearly 3 km/s. Time period is equal to that of earth rotation i.e., 24 hours. Polar Satellite Used for remote sensing, their orbit contains axis of rotation of earth. They can cover entire earth surface for viewing. BINARY STAR SYSTEM Two stars of mass M1 and M2 form a stable system when they move in circular orbit about their centre of mass under their mutual gravitational attraction. V2 M2 Applications If a body is projected with velocity greater than escape velocity, the intersteller speed is given by vI.S = If a body is projected upwards with velocity v = Kve, where K < 1, maximum height attained from centre R of earth ‘r’ = 1 − K 2 . RK 2 v 2R From surface of the earth h = r − R = 1 − K 2 = v 2 − v 2 If a body is dropped from rest form height h comparable to radius of earth, then maximum velocity with which it strikes the ground v = v e Time period of satellite is independent of mass of the satellite. Atmosphere on a planet is possible only if vrms < ve. Where vrms = rms speed of gas molecules. If angular velocity of earth is made 17 times, object placed at equator fly off. As we move from pole to equator, gravity decreases by 0.35%. If orbital velocity of any satellite very close to Earth surface increases by 41.4%, then the satellite will escape to infinity. If gravitation force ∝ rn, time period of a satellite ∝ r(1 – n)/2. ❑ ❑ ❑ I ∝r I ∝ 1 r 2 r = R