Gravitational field

About Gravitational field

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Gravitational FieldThis is any region or space around a mass in which the gravitational force of the mass is felt. The gravitational force is a force field which acts when two bodies are not in contact. Gravitational force between two massesGravitational force exists between two objects on the earth’s surface at a distance from each other. The pull of gravity acts on mass of all sizes. It is a vector quantity.Newton’s Law of GravitationThis law states that, the force of attraction between two bodies is directly proportional to the product of their masses and inversely proportional to the square of their separation (i.e their distance apart).Mathematically,Fα mM α 1/r2F = GmM/r2; where mM are the masses, r, is their distance apart, F is the force and G is the gravitational constant whose value is 6.67 x 10-11 Nm2 kg-2.Relationship between Acceleration due to gravity, g and the Gravitational constant, GThe earth is a sphere of radius, R, with mass, M concentrated at its centre. The distance of any object on the earth’s surface to the centre of the earth is R, the earth’s radius. The gravitational force of attraction of the earth on any mass, m, on the earth’s surface is given by: F = GmM/R2.This is the force of gravity on the mass due to the earth, i.e the weight of the object, mg, where g is the acceleration due to gravity. Thus, F = mg = GmM/R2 ----------------- (equation *)Hence the force per unit mass F/m = g, which means that g can be considered as the gravitational field strength. Its S.I units are (m/s2) and (N/kg). It is approximately constant at 9.8 Nkg-1 near the surface of the earth. From equation * above, g = GM/R2.This is the gravitational field strength due to the earth. Its value decreases with increasing distance fromits centre of mass.

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Acceleration due to Gravity on different PlanetsFor an object that is thrown upwards, the maximum height it reaches depends on:i) the initial velocity, uii) the acceleration due to gravity, gIf we consider an object which is thrown vertically upwards with an initial velocity, u, v2 = u2 - 2gh Now, at the instance the object reaches its maximum height, v = 0,Thus, we haveu2 = 2gh.If the objects are thrown upwards on different planets, but each time with the same initial speed, we can say,u2/2 = ghi.e gh = a constant. (u2/2) In other words, the height the object will reach is inversely proportional to the acceleration due to gravity. i.e g α 1/h So, when comparing different planetsg1h1 = g2h2 Therefore, g1/ h1 = g2/ h2Worked exampleA person can jump 1.5 m on the earth. How high could the person jump on a planet having twice the mass of the earth and twice the radius of the earth?SolutionLet mass of the earth = M;let the radius of the earth = RMass of the other planet = 2M;radius of the other planet = 2RLet the acceleration due to gravity on the earth and the other planet be g1 and g2 respectively.Height h1 reached by the person on earth = 1.5 mLet the height reached by the person on the other planet = h2Now, since g = GM/R2,

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Thereforefor earth, g1 = GM/R2 for the other planet, g2 = G.2M/(2R)2 = GM/2R2since g1/ h1 = g2/ h2, therefore g1/ g2 = h2/ h1g1/ g2 = (GM/R2)/(GM/2R2) = 2therefore, 2 = h2/ h1 , 2 = h2/ 1.5h2 = 2 x 1.5 = 3 mWeight of Objects in SpaceWhen an object is on or near the surface of the earth, it experiences the full extent of the force of gravity or acceleration due to gravity of the earth. The gravitational force of attraction between the object and the earth is given as:Case 1: When the object is on the surface of the earth F = GMm/R2Case 2: When the object is above the surface of the earth
F = GMm/(R+h)2Case 3: When the objects is below the surface of the earth
F = GMm/(R-h)2
mMR
R
mhMR
MmR

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Worked ExampleCalculate the force of attraction between the earth and an object of mass 20 kg if (i) the object is on the surface of the earth(ii) distance 30000m above the surface of the earth(iii) distance 30000m below the surface of the earth.[R = 6.4 x 106m, M = 5.97 x 1024 kg, G = 6.7 x 10-11 Nm2kg-2]Solution m = 20 kg, R = 6.4 x 106m, M = 5.97 x 1024 kg, F = ?(i) F = GMm/R2 = (6.7 x 10-11 x 5.97 x 1024 x 20)/( 6.4 x 106)2 = 195.3 N(ii) F = GMm/(R+h)2 = (6.7 x 10-11 x 5.97 x 1024 x 20)/( 6.4 x 106 + 30000)2 = 193.5 N(iii) F = GMm/(R-h)2 = (6.7 x 10-11 x 5.97 x 1024 x 20)/( 6.4 x 106 - 30000)2 = 197.2 NGravitational PotentialThe gravitational potential at a point in a gravitational field is defined as the workdone by an external agent in bringing a unit mass from infinity to that point. It is given as Vgrav = -GM/r (Jkg-1). Where M is the mass producing the gravitational field, r , is the distance apart and G is the gravitational constant.Gravitational potential, Vgrav, is the potential due to the gravitational field of the earth (or planet). Thus, it is also defined as the gravitational potential energy, (Egrav) given to an object per unit mass:Vgrav = - Egrav/m = - [GMm/R] / m = - GM / RGravitational Potential EnergyGravitational potential energy, Egrav, at a point in a gravitational field is defined as the workdone in bringing a mass, m, from infinity to that point.It is given as Egrav = GMm/r = m x Vgrav (J).For short changes in distance Δh which is far less than R (Δh<<R), gravitational potential energy is given by:Egrav ≈ m g ΔhWorked ExampleWhat is the gravitational potential at the Earth’s surface?

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[Mass of the Earth = 5.97 x 1024kg, Earth’s radius =6371 km, G = 6.7 x 10-11 Nm2kg-2]Solution Mass = 5.97 x 1024kg, R = 6.371 x 106m, G = 6.7 x 10-11 Nm2kg-2, Vgrav = ? Vgrav = - GM / R = - (6.7 x 10-11 x 5.97 x 1024) / 6.371 x 106 = - 6.278 x 107 J/kgEscape VelocityThis is the minimum speed needed for an object to “break free” from the gravitational attraction of a massive body such as the earth. That from the earth is about 40,270 km/h (11186.11 m/s).It is also the velocity at which the sum of an object’s kinetic energy and its gravitational potential energy equal to zero.That is:Gravitational Potential Energy + Kinetic Energy = 0Therefore, - [GMm/R] + ½ mv2 = 0½ mv2 = GMm/Rv2 =2GM/Rv = square root [2GM/R] ie The escape velocity can also be expressed in term of g.Since g =GM/R2 Note:1. At escape velocity, the object will move away forever from the massive body (such as earth), without additional acceleration applied to the object.2. As the object moves away from the massive body, the object will continually slow and (asymptotically) approach zero speed as the distance of the object approaches infinity.Worked exampleWhat is the escape velocity of a body launched from the earth’s surface? [g = 10 m/s2, R = 6.4 x 106 m]

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Solution.g = 10 m/s2, R = 6.4 x 106 m, Escape velocity, v = ?v = squareroot [2 g R] i.e v = (2 x 10 x 6.4 x 106)0.5 = 1.13 x 104 m/s.SatellitesBodies that orbit the larger and more massive bodies are regarded as satellites. The moon orbits round the Earth, Phobos and Deimos revolve round the Mars, these are satellites.Similarly, the Earth, Mars and other planets that orbit the Sun are regarded as the Sun’s satellites. Natural and Artificial SatellitesThe natural satellite is an orbiting body that occurs naturally in space. Eg moons that orbit planets are natural satellites.The artificial satellite is a man-made satellite that is made to orbit the planet (earth). Most artificial satellites revolve around the Earth. Their functions include the monitoring of the global weather, telecommunications, military and intelligence surveillance. Example of such is Hubble Space Telescope.