TJC 8 Gravitational Fields

Unit 8: Newton’s Law of Gravitation Learning Outcomes Candidates should be able to: (a) recall and use Newton's law of gravitation in the form F = 2 21 r mGm . (b) derive, from Newton's law of gravitation and the definition of gravitational field strength, the field strength due to a point mass, g = 2r GM . (c) recall and use g = 2r GM for the gravitational field strength due to a point mass to solve problems. (d) show an understanding that near the surface of the Earth , gravitational field strength is approximately constant and is equal to the acceleration of free fall. (e) define the gravitational potential at a point as the work done per unit mass by an external force in bringing a small test mass from infinity to the point. (f) solve problems using the equation  = r GM for the gravitational potential in the field due to a point mass. (g) show an understanding that the gravitational potential energy of a system of two point masses is UG = GMm r (h) recall that gravitational field strength at a point is equal to the negative potential gradient at that point and use this to solve problems. (i) analyse problems related to escape velocity by considering energy stores and transfers. (j) analyse circular orbits in inverse square law fields by relating the gravitational force to the centripetal acceleration it causes. (j) show an understanding of geostationary orbits and their application. Student’s copy

Temasek Junior College 2025 2 1. NEWTON’S LAW OF GRAVITATION LO(a) You may have heard the legend that Newton was struck on the head by a falling apple while napping under a tree. This alleged accident supposedly prompted him to imagine that perhaps all objects in the Universe were attracted to each other in the same way the apple was attracted to the Earth. It was in 1687 that Newton first published his work on the law of gravitation. Gravitational force F = 2 21 r mGm where m1 and m2 are the two point masses and r the distance between them. The constant of proportionality G is called the universal gravitational constant. G = 6.67  10-11 N m2 kg-2. Note: 1. The expression is applicable ONLY between point masses or spheres of uniform density. 2. For spheres, r is the distance between their centres and NOT surface to surface. 3. The gravitational force is directed along the line joining their centres. 4. The gravitational forces acting on the two masses form an action-reaction pair. 5. By the usual convention in Physics, an attractive force should have a negative sign (i.e. F = - 2 21 r mGm ). But since gravitational repulsion does not exist, the negative sign is omitted. 6. Newton’s law of gravitation is an example of an inverse square law of force. That is, the force varies inversely with the square of the distance between the two masses. 7. The value of G (= 6.67  10-11 N m2 kg-2) is very small. Thus, although gravitational force of attraction exists between all bodies, it is a weak force and is only significant when very massive objects are involved. Newton's law of gravitation states that two point masses attract each other with a force that is proportional to the product of their masses and inversely proportional to the square of their separation. r m1 m2 F F

Temasek Junior College 2025 3 Example 1: Find the gravitational force which (a) a person of mass 60 kg exerts on another person of the same mass separated by a distance of 1.0 m. Solution: (b) the Earth of mass 6.0  1024 kg exerts on a man of mass 60 kg standing at its surface. The Earth has a radius of 6.4  106 m. Solution: 2. THE GRAVITATIONAL FIELD LO(b) Any mass M sets up a “field of force” or gravitational field around it. This field is invisible but it can be detected by placing a small point mass m nearby. This small mass m, also called a test mass, experiences an attractive gravitational force F. Gravitational field strength g = gravitational force per unit mass = m F Unit of g: N kg-1. It is a vector quantity. The direction of the gravitational field strength is the same as the direction of the force that acts on a small mass placed in the field. Therefore g points in the direction towards the mass M. The gravitational field is a region of space where a mass experiences a gravitation force. The gravitational field strength at a point is defined as the gravitational force acting per unit mass at the point.

Temasek Junior College 2025 4 Exercise: Show, using base units, that N kg-1 is equivalent to m s-2. Solution: 2- -1 -2-1 s m = kg s m kg = kg N At any point, the gravitational field strength is equal to the acceleration due to gravity. 2.1 Derivation of gravitational field strength g due to a point mass M LO(b),(c) From Newton’s law of gravitation, the attractive gravitational force on a point mass m caused by another point mass M, with a distance r between their centres, is given by By definition, the gravitational field strength due to the mass M at a distance of r from its centre is thus given by Since g  2r 1 , the field strength obeys the inverse square law with distance. Thus the gravitational field is sometimes known as an inverse square law field. Example 2: The International Space Station (ISS) operates at an altitude h of 350 km. What is the gravitational field strength at its orbit? Given: Mass of Earth ME = 5.98  1024 kg, Radius of Earth RE = 6370 km. Solution: r GM=m F=g 2 r GMm=F 2 x M r m g = m F = 2r GM

Temasek Junior College 2025 5 Example 3: Figure shows how the gravitational field strength g varies with distance r from the centre of a planet of radius 2.7 x 107 m. Show that the field strength obeys an inverse square law. 2.2 Variation of g with distance r from the Earth's centre Outside the Earth Outside the Earth, assuming that it is spherical, the Earth behaves as a point mass, with all the mass of the sphere concentrated at its centre. Any small mass m outside the Earth is attracted by the total mass M of the Earth. Hence the gravitational field strength of the Earth for r > RE is given by equation Inside the Earth (assuming uniform density) Inside the Earth, a small mass m at any distance r1 , is attracted by the mass M1 of the shaded region of radius r1. Graph of variation of g from inside Earth to outside Earth M m g Solution: Choose any 3 points on the curve and show that gr2 = constant r/107 m g/Nkg-1 gr2/ 107 Nkg-1m2 4.0 10.0 160 6.0 4.5 160 8.0 2.5 160 Since gr2 = constant,  g = 2r 1 R E  g  2r 1 r GM g 2 = 2 1 1 r MG=g 1 2 1 r ρVG= 1 1 2 1 rGρ3 4=r r3 4ρ G= 3    g  r M1 r1 9.81 m Inside Earth g  r Outside Earth g  1/r2 RE 0 r1 Inside the Earth, at a distance r1 from centre, only the mass M1 of the shaded region contributes to g. r g r1 M1

Temasek Junior College 2025 6 2.3 Gravitational Field Lines