r^{2}[/t][/t] or, the gravitational force of between two point masses is proportional to the product of their masses & inversely proportional to the square of their separation. Gravitational field strength at a point is the gravitational force per unit mass at that point. It is a vector and its S.I. unit is N kg^{1}. By definition , g = F / m By Newton Law of Gravitation, F = GMm / r^{2} Combining, magnitude of g = GM / r^{2} Therefore g = GM / r^{2}, M = Mass of object “creating” the field Example 1: Assuming that the Earth is a uniform sphere of radius 6.4 x 10^{6} m and mass 6.0 x 10^{24} kg, find the gravitational field strength g at a point: (a) on the surface, g = GM / r^{2} = (6.67 × 10^{11})(6.0 x 10^{24}) / (6.4 x 10^{6})^{2} = 9.77ms^{2} (b) at height 0.50 times the radius of above the Earth's surface. g = GM / r^{2} = (6.67 × 10^{11})(6.0 x 10^{24}) / ( (1.5 × 6.4 x 10^{6})2 = 4.34ms^{2} Example 2: The acceleration due to gravity at the Earth's surface is 9.80ms^{2}. Calculate the acceleration due to gravity on a planet which has the same density but twice the radius of Earth. g = GM / r^{2} gP / gE = M_{P}r_{E}^{2} / M_{E}r_{P}^{2} = (4/3) π r_{P}^{3}r_{E}^{2}ρ_{P} / (4/3) π r_{E}^{3}r_{P}^{2}ρ_{E} = r_{P} / r_{E} = 2 Hence g_{P} = 2 x 9.81 = 19.6ms^{2} Assuming that Earth is a uniform sphere of mass M. The magnitude of the gravitational force from Earth on a particle of mass m, located outside Earth a distance r from the centre of the Earth is F = GMm / r^{2}. When a particle is released, it will fall towards the centre of the Earth, as a result of the gravitational force with an acceleration a_{g}.
F_{G} = ma_{g} a_{g} = GM / r^{2} Hence a_{g} = g Thus gravitational field strength g is also numerically equal to the acceleration of free fall. Example 1: A ship is at rest on the Earth's equator. Assuming the earth to be a perfect sphere of radius R and the acceleration due to gravity at the poles is g_{o}, express its apparent weight, N, of a body of mass m in terms of m, g_{o}, R and T (the period of the earth's rotation about its axis, which is one day). At the North Pole, the gravitational attraction is F = G M_{E}m / R^{2} = mg^{o} At the equator, Normal Reaction Force on ship by Earth = Gravitational attraction  centripetal force N = mg_{o} – mRω^{2}= mg_{o} – mR (2π / T)^{2} Gravitational potential at a point is defined as the work done (by an external agent) in bringing a unit mass from infinity to that point (without changing its kinetic energy).
φ = W / m = GM / r Why gravitational potential values are always negative? As the gravitational force on the mass is attractive, the work done by an ext agent in bringing unit mass from infinity to any point in the field will be negative work {as the force exerted by the ext agent is opposite in direction to the displacement to ensure that ΔKE = 0} Hence by the definition of negative work, all values of φ are negative. g = [/t][/c][/t] dφ =  gradient of φr graph {Analogy: E = dV/dx} dr[/t][/t] Gravitational potential energy U of a mass m at a point in the gravitational field of another mass M, is the work done in bringing that mass m {NOT: unit mass, or a mass} from infinity to that point.
→ U = m φ = GMm / r Change in GPE, ΔU = mgh only if g is constant over the distance h; {→ h<< radius of planet} otherwise, must use: ΔU = mφ_{f}mφ_{i}  Aspects  Electric Field  Gravitational Field  1.  Quantity interacting with or producing the field  Charge Q  Mass M  2.  Definition of Field Strength  Force per unit positive charge E = F / q  Force per unit mass g = F / M  3.  Force between two Point Charges or Masses  Coulomb's Law: F_{e} = Q_{1}Q_{2} / 4πε_{o}r^{2}  Newton's Law of Gravitation: F_{g} = G (GMm / r^{2})  4.  Field Strength of isolated Point Charge or Mass  E = Q / 4πε_{o}r^{2}  g = G (GM / r^{2})  5.  Definition of Potential  Work done in bringing a unit positive charge from infinity to the point; V = W /Q  Work done in bringing a unit mass from infinity to the point; φ = W / M  6.  Potential of isolated Point Charge or Mass  V = Q / 4πε_{o}r  φ = G (M / r)  7.  Change in Potential Energy  ΔU = q ΔV  ΔU = m Δφ  Total Energy of a Satellite = GPE + KE = (GMm / r) + ½(GMm / r) Escape Speed of a Satellite By Conservation of Energy, Initial KE  +  Initial GPE  =  Final KE  +  Final GPE  (½mv_{E}^{2})  +  (GMm / r)  =  (0)  +  (0)  Thus escape speed, v_{E} = √(2GM / R) Note : Escape speed of an object is independent of its mass For a satellite in circular orbit, "the centripetal force is provided by the gravitational force" {Must always state what force is providing the centripetal force before following eqn is used!} Hence GMm / r^{2} = mv^{2} / r = mrω^{2} = mr (2π / T)^{2} A satellite does not move in the direction of the gravitational force {ie it stays in its circular orbit} because: the gravitational force exerted by the Earth on the satellite is just sufficient to cause the centripetal acceleration but not enough to also pull it down towards the Earth. {This explains also why the Moon does not fall towards the Earth} Geostationary satellite is one which is always above a certain point on the Earth (as the Earth rotates about its axis.) For a geostationary orbit: T = 24 hrs, orbital radius (& height) are fixed values from the centre of the Earth, ang velocity w is also a fixed value; rotates fr west to east. However, the mass of the satellite is NOT a particular value & hence the ke, gpe, & the centripetal force are also not fixed values {ie their values depend on the mass of the geostationary satellite.} A geostationary orbit must lie in the equatorial plane of the earth because it must accelerate in a plane where the centre of Earth lies since the net force exerted on the satellite is the Earth's gravitational force, which is directed towards the centre of Earth. {Alternatively, may explain by showing why it's impossible for a satellite in a nonequatorial plane to be geostationary.} 



