(a) (i) State Newton’s Law of Universal Gravitation.
(ii) Define gravitational field.
(b) (i) Derive the equation relating the universal gravitational constant, G, and the acceleration of free fall, g, at the surface of the earth from Newton’s law of universal gravitation.
(ii) State two assumptions for which the relationship in 8(b)(i) holds.
(c) Calculate the force of attraction between a star of mass 2.00 x 1030 kg and the earth assuming the star is located 1.50 x 108 km from the earth. [Mass of the earth = 5.98 x 1024kg; G = 6.67 x 10-11N m\(^{2}\) kg-2; g = 10 m s\(^{-2}\)
(d) (i) Define escape velocity.
(ii) State two differences between the acceleration of free fall (g) and the universal gravitational constant (G).
(a)(i) Newton's Law of Universal Gravitation: Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres.
\[ F = \frac{G m_1 m_2}{r^{2}} \]
(a)(ii) Gravitational field: A gravitational field is a region of space in which a body of mass experiences a force of attraction. Its strength at a point is the force per unit mass acting on a small mass placed at that point.
(b)(i) Relationship between G and g: Consider a body of mass \(m\) resting on the earth's surface. The earth (mass \(M\), radius \(R\)) attracts it with a force which, by Newton's law, is
\[ F = \frac{G M m}{R^{2}}. \]
This same force is the weight of the body, \(F = mg\). Equating the two:
\[ mg = \frac{G M m}{R^{2}} \quad\Rightarrow\quad g = \frac{G M}{R^{2}}. \]
(b)(ii) Assumptions: (1) The earth is a perfect sphere of uniform density, so its whole mass may be taken to act at its centre. (2) The body is small compared with the earth, and effects such as the earth's rotation and air resistance are neglected.
(c) Force between the star and the earth:
\[ r = 1.50\times10^{8}\,\text{km} = 1.50\times10^{11}\,\text{m} \]
\[ F = \frac{G M_{star} M_{earth}}{r^{2}} = \frac{(6.67\times10^{-11})(2.00\times10^{30})(5.98\times10^{24})}{(1.50\times10^{11})^{2}} \]
\[ F = \frac{7.98\times10^{44}}{2.25\times10^{22}} \approx 3.55\times10^{22}\,\text{N}. \]
(d)(i) Escape velocity: The minimum velocity with which a body must be projected from the surface of the earth (or a planet) so that it completely overcomes the gravitational pull and escapes without ever returning. \(v_e = \sqrt{2gR}\).
(d)(ii) Two differences between g and G:
- g (acceleration of free fall) varies from place to place on the earth and from planet to planet, whereas G (universal gravitational constant) has the same value everywhere in the universe.
- g is a vector with unit m s-2, while G is a scalar constant with unit N m2 kg-2.
