1 Newton s Gravitational Law Gravity exists because bodies have masses. Newton s Gravitational Law states that the force of attraction between two point masses is directly proportional to the product of the mass and inversely proportional to the square of the distance between the masses. This leads to the relationship F m 1m 2 By inserting the universal Gravitational constant, G, we get that the equation of Newton s Gravitational Law: F = G m 1m 2 where G = 6.67 10 11 Nm 2 kg 2 Conditions If the two bodies have spherical symmetry (eg. the Sun and the Earth) If one body has a spherical symmetry and the other is small compared with the separation of their centers (eg. the Earth and a brick) Both the bodies are not spherically symmetrical but both are small compared with the separation of their centers (eg. two bricks that are a few centimeters apart) Kepler s Law These three laws describe the planetary motion: 1. All planets move in elliptical orbits with the sun at the focus. 2. Each planet moves in such a way that the imaginary line joining the Sun sweeps out equal areas in equal time. 3. The square of the period of a revolution of any planet about the sun us directly proportional to the cube of the mean of the distance from the Sun. (T 2 r 3 )
2 From the diagram below, the average speed from A to B is greater than the average speed from C to D in such a way that: Area of ABS = Area of CDS Showing that Kelper s Third Law is Consistent with Newton s Gravitational Law Consider a planet of mass m, moving about the Sun in a circular orbit of radius r. The angular speed of the planet is ω and the mass of the Sun is m s. Using Newton s 2 nd Law of Motion (F = ma) we get that: F = G m sm eq i r2 F = m ω 2 r eq ii Equating equation i to equation ii we get that: G m sm = m ω2 r
3 We also know that: ω = 2π T where T is the period for one revoultion of the planet Therefore G m sm = m (2π T ) 2 r G m sm 4π2 = m r r2 T 2 T 2 = m r 4π2 G m s m T 2 = 4π 2 r 3 G m s From this equation we can observe that indeed T 2 r 3 This is Kepler s 3 rd Law, which is derived using Newton s Gravitational Law. Therefore the two laws are consistent. Gravitational Field Strength Gravitational field strength (g) at a point in a gravitational field is defined as a force per unit acting on a mass (m) placed at a point. g = F m Gravitational Field Strength Due to a Point Mass The force on a point mass (m) at a distance (r) from the point mass (M) is given by: F = G m M
4 Since g = F, therefore F = mg, we get that m mg = G mg = G m M m M g = G m M m g = GM This equation can also be applied outside an extended object of mass (M) at a distance (r) from its centre provided the object is spherically symmetrical and the distance from its centre to its surface is small compared with r. (NB: If M E represents the mass of the Earth and represents the radius of the Earth, the field on the surface is given as g = GM E 2 = 9.81 Nkg 1 Diagram A below shows the gravitational field of the Earth
5 Diagram B below shows the Earth s gravitational field over a small area. Gravitational Potential and Potential Energy The gravitational potential at a point in a gravitational field is defined as being the work done per unit mass in bringing a mass (m) from infinity (where the potential is considered to be zero) to a point. U = W m where U is gravitational potential at some point and W is the work done in bringing mass of infinity to that point. It follows from the equation above that in general: U = G m r Where U is the gravitational potential due to a body of mass (m) at a point outside of the body and distance (r) from the centre. Assuming that the body is symmetrical and is small compared to r. The Gravitational Potential Energy (PE) of the body of mass (m) at a point where the gravitational potential is U can be given by: PE = mu (NB: the Gravitational Potential due to the surface of the Earth (U E ) U E = GM E
6 Escape Velocity If a ball is thrown upwards from the Earth s surface, its speed decreases from the moment it is released due to the Earth s gravitational field. The height the ball reaches depends on the speed with which it is projected (the greater the speed the greater the height). In order for the ball to escape from the Earth, it must be projected with a velocity which is at least great enough for the ball to reach infinity before coming to rest. The minimum velocity that achieves this is known as the escape velocity. Let us consider a body of mass(m) which is projected upwards with a velocity (v) from the Earth s surface. When the body reaches a distance (r), it feels a gravitational attraction (F) due to the Earth, which can be calculated using the equation: F = GmM E 2 The work done (W) by the body in moving from the Earth s surface to infinity is given by: W = GmM E In order for the body to do this work and escape, it must have the same amount of kinetic energy at the moment of projection. Hence 1 2 mv2 = GmM E v 2 v 2 = 2 GmM E m = 2 GM E v = 2 GM E This calculation only applies to bodies that are not driven or powered flights. The escape velocity does not depend on the direction of the projection. This is because the kinetic energy lost by the body only depends on the height it reaches and not the path taken to reach the height. (NB: The estimated value of the escape velocity on Earth is v 11 km /s )
7 Satellite Motion and Orbits Satellites are objects which are in orbit around a larger mass as a direct result of gravitational attraction. Our planets are satellites of the Sun and the Moon is a satellite of the Earth. Artificial Earth satellites behave the same way that the Moon does. A satellite orbit around the Earth has a centripetal acceleration equal to the free- fall acceleration at the particular position of the satellite. A geostationary satellite is one that stays over one location on the Earth s surface all the time. The satellite must be in orbit with the correct radius such that its orbital period is the same as the Earth s rotation. The satellites must be in orbit over the equator. The radius of a geostationary satellite is given as: r 3 = GM E T 2 4π 2 Hence if calculated r 4.24 10 7 m Example 1 Calculate the gravitational attraction force between the bodies of mass 3 kg and 2 kg, placed with their centres 50 cm apart. (Assume that G = 6.67 10 11 Nm 2 kg 2 ) Example 2 Two particles of mass 0.2 kg and 0.3 kg are placed 15 cm apart. A third mass of 0.05 kg is placed between them on a line joining the first two particles. Calculate: a) the gravitational force acting on the third particle if it is placed 5 cm from the 0.3 kg mass. b) Where along the the line must the third particle be placed for no gravitational force to act on it. (Assume that G = 6.67 10 11 Nm 2 kg 2 )
8 Example 3 Three isolated uniform spheres of mass 3 kg, 4 kg and 8 kg are placed at the corners of right angle triangle as shown in the diagram below. Calculate the net gravitational force on the 4 kg sphere. 3kg 3.0m 4 kg 8kg 4.0m Example 4 Assuming that the Earth is a uniform sphere of radius 6.4 10 6 m and mass of 6.0 10 24 kg, calculate the gravitational strength (g) at a point on the surface of the Earth. Example 5 Find the gravitational field intensity (g) on the surface of the Jupiter, given that the radius of Jupiter is 6.99 10 7 m and its mass is 1.90 10 27 kg. Example 6 Given that the mean radius of the Moon in its orbit is 1.74 10 6 m, find the gravitational potential energy of the moon in its orbit. If the masses on the Earth and the Moon are 5.98 10 24 kg and 7.36 10 22 kg respectively.
9 Example 7 Calculate the minimum speed which a body must have to escape from the Moon s gravitational field. Given that the Moon has a mass of 7.36 10 22 kg and a radius of 1.7 10 6 m. Example 8 Satellites which orbit the Earth with a time period of 24 hours are used for communication purposes since they appear to be stationary above a given point on the Earth. Calculate the height of such a satellite above the Earth s surface. Assuming that the mass of the Earth is 6.0 10 24 kg and the radius is 6.4 10 6 m.