JPJC 2026 Gravitational Field Lecture Notes Tutor

1 JURONG PIONEER JUNIOR COLLEGE 9478 H2 PHYSICS Gravitational Fields Introduction  Isaac Newton ( 1642 –1726), published Philosophiæ Naturalis Principia Mathematica ("Mathematical Principles of Natural Philosophy") in 1687. In it, he formulated the Newton’s laws of motion (Dynamics), and the Newton’s law of gravitation.  While the popular story involves Newton observing a falling apple, his actual discovery was more complex. Newton built upon earlier work, including Johannes Kepler's mathematical descriptions of planetary motions from the early 17th century. The law of gravity is celebrated for its elegant simplicity and broad applicability. The applications range from understanding planetary orbits, satellite motion, tidal flow, galaxy formation and stellar dynamics. This law, along with Newton's laws of motion, provided a foundation for understanding the mechanics of the universe. 1 GRAVITATIONAL FORCE 1.1 Newton’s law of gravitation Newton’s law of gravitation states that the force of attraction between two point masses is directly proportional to the product of the masses and inversely proportional to the square of the distance between them.  Point Mass: a point mass is a theoretical concept representing an object with a finite mass that occupies zero space, meaning it has no physical dimensions. 1m 2m Fig. 1 Gravitational force between two point masses  Mathematically, it is written as 1 2 2g m mF G r where G is the gravitational constant with a value of 11 2 26.67 10 N m kg  , and are the masses of the two bodies respectively, r is the distance between their centres of masses of these 2 point masses, and Fg is the magnitude of the gravitational force of attraction between them. 1m 2m r 1 2 2 m mF G r (a) Recall and use Newton’s law of gravitation in the form .

2  This is known as an inverse square law, as the magnitude of the force varies with the inverse square of the separation of the particles, 2 1gF r Fig. 2 Magnitude of gravitational force variation with separation  The gravitational forces of attraction between two bodies form an action-reaction pair, consistent with Newton’s third law. Mass m1 attracts m2 with the same magnitude of force as m2 attracts m1, but in opposite directions on each mass.  For spherical masses, the distance r is measured between their respective centres of mass (assume the mass has uniform density and concentrated at its centre).  Sometimes the equation is written with a ‘─’ sign as 1 2 2 m mF G r  to indicate the attractive nature of the force, where r is positive when pointing way from the mass. The negative sign is normally ignored when only magnitude is required. If direction is to be considered, the F-r graph is plotted on the negative side. Fig. 3 Gravitational force (including direction) with separation  Because G is very small in value (6.67x10 -11), the gravitational force is often ignored unless one of the masses is sufficiently large like the Moon (7.3 x 1022 kg), the Earth (6.0 x 1024 kg) or the Sun (2.0 x 1030 kg). E.g. we can ignore the negligible gravitational force of attraction between 2 human beings.

3 Example 1 A man of mass 85.0 kg is standing on the surface of the Earth. Given that the mass of the Earth is 5.97 × 1024 kg with a radius of 6.37 × 106 m, calculate the gravitational force the Earth exerts on the man. Hence, state the gravitational force the man exerts on the Earth. Solution: (Diagram not to scale) 2 24 11 6 2 (5.97x10 )(85.0)(6.67x10 ) (6.37x10 ) 834 N g MmF G r     This is equal to the weight, mg = 85 x 9.81 = 834 N The gravitational force that the Earth exerts on the man is 834 N towards the centre of Earth. The gravitational force that the man exerts on the Earth is 834 N from centre of Earth towards the man. Example 2 Two stationary objects of masses 35.0×10 kg and 3101.0 kg are placed at a distance of 10 m apart. A third object of mass 1.0 kg, placed between the two masses, experiences no resultant force. (a) On Fig. 1.1, draw the forces acting on the 1.0 kg object. (b) Calculate the distance x between the 1.0 kg and the 1000 kg objects. Solution: (a) Fig. 1.1 (b) Since the 1.0 kg object experiences zero resultant force,   2 32 2 31 10 x MMG x MMG     22 1000 10 5000 xx     5000 1000 10 2 2   x x Taking square root on both sides of the equation, 10 m kg x kg .0 kg

4   5 1 10  x x xx 510  1.3x m Therefore, the distance x between the 1.0 kg and 1000 kg objects is 3.1 m. 2 GRAVITATIONAL FIELD 2.1 Concept of a gravitational field How does the Earth exert an attractive force on the Moon over a vast empty space without making any contact with the Moon? Physicists use the concept of a field as an intermediary of interaction between masses. The Earth as a mass produces a gravitational field at every point in the space from itself to infinity. Any other mass placed in this field will experience an attractive force. The force acting on the Moon therefore arises from interaction of the gravitational field of the Earth and the mass of the Moon.  A gravitational field is a region of space in which a mass experiences a force due to the presence of another mass (which generates the gravitational field).  At every point around a mass, there is a vector- value gravitational field which has both magnitude and direction. This gravitational field can be represented by field lines with vector direction, directed towards the centre of the mass. The relative strength of the field is represented by separation of the field lines. Closer field lines mean stronger field strength. The field direction follows the direction of the gravitational force acting on other masses. The field lines do not cross each other. Fig. 4 Gravitational field lines of a mass [from Energy and Fields] Candidates should be able to (f) show an understanding of the concept of a field as a region of space in which bodies may experience a force associated with the field. (h) represent gravitational fields by means of field lines (g) define gravitational field strength at a point as the gravitational force per unit mass on a mass placed at that point (i) show an understanding that the force on a mass in a gravitational field acts along the field lines (b) derive, from Newton’s law of gravitation and the definition of gravitational field 2 Mg G r strength, the field strength due to a point mass,

5 2.2 Gravitational field strength To measure the gravitational field strength produced by M at a point X, at distance r from M, we placed a small test mass m at point X. (The small test mass is used to test or measure the force without affecting M). The gravitational force acting on m by M, divided by m is the gravitational field strength produced by M at X. M is the source of the field, m is used to test the field strength produced by M. Fig. 5 Gravitational field strength at point X due to an object of mass M  Units for g is N kg−1, from the definition.  If 2 or more fields are present at a point the resultant gravitational field strength is obtained by vector addition. The direction of the resultant gravitational force is along the direction of the resultant field lines with a magnitude of F = mgresultant. Fig. 6 Vector addition of gravitational field strength Example 3 Using Newton’s law of gravitation and the definition of gravitational field strength, derive the equation 2r GMg  . Solution: Consider a mass M and another mass m placed at point X, which is at a distance r away from mass M. gFg m M m r point X M point X r m Gravitational field strength at a point is defined as the gravitational force per unit mass acting at that point.

6 By Newton’s law of gravitation, the force gF experienced by mass m is 2r GMm . From the definition of gravitational field strength at point X,  gF mg m experienced by massat point X mass 2