HCI 07A Gravitation Solutions (Discussion Questions)

Gravitation Tutorial 7A Discussion Questions Suggested Solutions D1 F 100 = F400 O R s i m p l y g100 = g400 𝐺 ሺଵ଴଴ሻ௠ ௫మ ൌ𝐺 ሺସ଴଴ሻ௠ ሺଵ଴.଴ି௫ሻమ 𝐺 ሺଵ଴଴ሻ ௫మ ൌ𝐺 ሺସ଴଴ሻ ሺଵ଴.଴ି௫ሻమ x = 3.33 m D2 B Vector sum of gravitational field strengths at B due to each of the 3 bodies = 0 at the neutral point Gravitational field strength at a point due to a mass is propor tional to the mass and inversely proportional to the square of the distance between the mass and the point. Consider Earth and Moon’s intera ction first, the point where th e resultant gravitational field strength is zero is nearer to the smaller mass (refer to D1 whe re g = 0 is nearer the 100 kg mass). Then consider Moon and Sun’s interaction, which the neut ral point will be shifted nearer to the Sun, but it will be still nearer to the smaller mass. Good to know:   Mass of the Earth = 5.97 x 10 24 kg,  mass of the Moon = 7.35 x 1022 kg,   mass of the Sun = 1.99 x 1030 kg  D3 2 2 98 1E E E MgG . m sR  (a) 2 22 1 new E EE n e w g Rg Rg R  2 2 2 1 24 524 E new E E E Rgg g . m s(R)   m  F400 F100 x 10.0 - x 100 kg  400 kg  NOTE: The concept of finding the point where the resultant force  acting on a mass due to the two masses equals zero is equivalent  to finding the point where the resultant gravitational field  strength due to the two masses equals zero.  The concept of using ratio is a common method to  solve many physics problems.

(b)     3 22 4 43 3 GRGMgG R RR gR     Same radius but twice the density  g’ = 19.6 m s-2 (c) Half radius but twice the density  g’ = 9.81 m s-2 D4 B 3 22 4 43 3 65 1 8 3.613 5 EE E E E MM M M M GrGMgG r g r rr gr r r gr r r                D5 Apparent weight of zero means no force on its support  normal force by ground on it = 0 The entire gravitational force provides the centripetal force.  cWNm a (since N = 0)    2 2 cWm a mg mr T Taking g at the surface to be 9.81 m s-2, and the given radius of Earth, T = 5100 s (2 s.f.) D6 (2009 P3 Q5) (a)(i) The gravitational field strength at a point is the gravitational force per unit mass acting on a small test mass placed at that point. (ii) Newton’s Law of Gravitation states that every point mass (or particle) attracts every other point mass (or particle) with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between t hem. Thus 2 GMmF R 2 2 Since , GMmmg R GMg m R Fg   (b)(i) Average density =  30 17 3 34 total mass 5.2 10 2.53 104total volume 1.7 103 kg m     The concept of using ratio again similar to D3.

(ii) Gravitational force acts toward the centre of the neutron star; all the particles are pulled inwards. The outer layers compress the inner layers, resulting in increased density towards the centre. (c)(i)     11 30 12 2 22 4 6.67 10 5.2 10 1.20 10 1.7 10 g GM msR     (ii) a = r2 = (1.7 x 104) 2 2 0.21   = 1.52 x 107 m s-2 (iii) Suppose there is normal contact force N acting on the particle by the surface of the star. Applying Newton’s 2 nd Law, centripetal force = gravitational force – normal contact force, i.e. ma = mg – N, Based on the computed values of a and g, N = m (g – a) > 0. Thus the particle does not leave the surface, but remain in con tact with the surface. The gravitational force is more than enough to provide the required centripetal force.     D7 CIE J96/III/2(part) (c)(i)     11 24 22 3 6.67 10 5.98 10 (1.00) 9.83 6370 10 g GMm NF R    (ii) 2 2 3 2 4(1.00)(6370 10 ) 0.0337( 2 46 06 0 ) c NFm r      (iii) Fg – T = Fc = 0.0337 T = 9.80 N (d)(i) 9.83 m s -2 (ii) 9.80 m s -2 (e) The acceleration due to gravity is actually larger than the measured acceleration by the amount of centripetal acceleration due to the Earth’s rotation . What the student measured should be called the acceleration of free fall. D8 2018 P1 Q11 C Without firing its rocket, gravity or gravitational force is the only force acting on the spacecraft. Under the influence of gravity alone, it is possible for paths A, B (spacecraft in circular motion) and D (spacecraft projected away from Earth).

D9 2021 P2 Q4(b) Gravitational force provides for centripetal force. 2 2 GMm mrr  32GM r     2 11 24 3 26 6 71 0 6 01 0 110 60.. r      67 615 10 mr.     11 24 -1 22 6 6 6 71 0 6 01 0 69 N k g 7 615 10 ..GMg. r .     D10 2017 P2 Q2 (Part) 2 (a)(i)  72 1 75 102 549778 549778 s 6 3632 = 6.36 days200 24 3600 .rT. v        Note: the distance and velocity must be converted to S.I. units. Since this is a “show” question, students should express the an swer in terms of seconds first then show the steps to convert to days. Should also show that they have hit the calculator by writing the more exact value of 6.3632. 2 (a)(ii) 1. Charon will be above the same place on the surface of Pluto at all times. 2 (a)(ii) 2. The same face of Charon will be seen from Pluto at all times. 2 (b)     11 22 1 22 6 6 67 10 1 31 10 0 607 N kg 12 0 1 0 Pluto P Pluto ..GMg. r .     

D11 2016 P3 Q9 (part) (i) The gravitational forces of attraction between the two stars provide the centripetal forces necessary for the stars to go in circular motion. This pair for forces constitute a Newton’s 3rd law action-reaction pair, which are always equal in magnitude (and opposite in direction). (ii) Assuming it is Earth year, 𝜔ൌ ଶగ ் ൌ ଶగ ସ.଴ൈଷ଺ହൈଶସൈ଺଴ൈ଺଴ ൌ4 . 9 8ൈ1 0ି଼ rad s-1 (iii) Since the centripetal forces are equal in magnitude, M ArA2 = MBrB2 Since they also have common , rB = d – rA , we can re-arrange the above equation: 𝑟஺ ൌ 𝑀஻ 𝑀஺ ሺ𝑑െ𝑟 ஺ሻ 𝑀஺ 𝑀஻ ൌ ሺ𝑑െ𝑟 ஺ሻ 𝑟஺ ൌ 𝑑 𝑟஺ െ1 𝑀஺ 𝑀஻ ൅1ൌ 𝑑 𝑟஺ 𝑟஺ ൌ ௗ ൬ಾಲ ಾಳ ାଵ൰ ൌ ଷ.଴ൈଵ଴భభ ሺଷ.଴ାଵሻ ൌ 7.5 x 1010 m (iv) By Newton’s 2 nd Law, Fnet = ma; 𝐺𝑀஺𝑀஻ 𝑑ଶ ൌ𝑀 ஺𝑟஺𝜔ଶ 𝑀஻ ൌ 𝑑ଶ𝑟஺𝜔ଶ 𝐺 ൌ ሺ3.0 ൈ 10ଵଵሻଶሺ7.5 ൈ 10ଵ଴ሻሺ4.98 ൈ 10ି଼ሻଶ 6.67 ൈ 10ିଵଵ MB = 2.51 x 1029 kg Since ெಲ ெಳ ൌ3 . 0, MA = 3.0 (2.51 x 1029) = 7.53 x 1029 kg Comments: students should distinguish clearly between the symbols for quantities related to each star. (v) In each period, there will be two instants when the two stars form a straight line with Earth. At these instances, the intensity will dip because one star is behind the other. Therefore, the fluctuation in intensity is half the period of the orbit of the stars.