YEAR 12 PHYSICS: GRAVITATION PAST EXAM QUESTIONS Name: QUESTION 1 (1995 EXAM) (a) State Newton s Univesal Law of Gavitation in wods Between any two masses, thee exists a mutual attactive foce. This foce is diectly popotional to the poduct of the masses and invesely popotional to the squae of the distance between them. This foce acts along a line between the centes of the masses. (b) A satellite of mass (m) moves in obit of a planet with mass (M). The satellite, m, is smalle in mass than the planet, M. Assume that the satellite moves aound the planet in a cicula obit with a adius, R with a constant speed, v. (i) Explain how it is possible fo the gavitational foce to cause the satellite to acceleate while its speed emains constant. Its speed maybe constant in magnitude but the diection of this motion is constantly changing ove time due to the gavitational foce constantly pulling the satellite towads the cental mass. Since this foce is constant and the speed is constantly changing pe unit time, the body is acceleation.
(ii) If T is the peiod of the satellite in its obit aound the planet, show that the adius of the 3 obit of the satellite is = Gm 2T 2 given that T 2 = 4π2 3 4π 2 Gm 2 T 2 = 4π2 3 Gm 2 3 = Gm 2T 2 4π 2 3 = Gm 2T 2 4π 2 (iii) Calculate the moon s obital adius given that its peiod of otation is appoximately 27 days and 7 hous (2.3582 x 10 6 s) 3 = Gm 2T 2 4π 2 3 = 6.67 x 10 11 x 5.977 x 10 24 x (2.3582 x 10 6 ) 2 4π 2 3 = 2.217 x 1027 4π 2 = 3. 83 x 10 8 m
QUESTION 2 (1996 EXAM) Whilst obiting the Eath, the space shuttle Endeavou had a velocity of 7.8 x 10 3 ms -1 (a) Calculate the adius of its cicula obit V = GM = GM V 2 = 6. 67 x 10 11 x 5. 97 x 10 24 (7. 8 x 10 3 ) 2 = 6. 54 x 10 6 m (b) Two isolated masses M and m ae sepaated by a distance,. The mass M is twice the mass of the smalle body, m. On the diagam above, daw vecto aows to illustate the gavitational foce (F) on each mass HINT: Despite one mass being lage, the gavitational foce acting on each object is equal in magnitude but opposite in diection.
QUESTION 3 (1999 EXAM) The unifom cicula motion of a space vehicle in a cicula obit ound a planet is caused by the gavitational foce between the planet and the vehicle. (a) Calculate the magnitude and diection of the gavitational foce on a space vehicle of mass m= 1.00 x 10 3 kg at a distance of 1.65 x 10 7 m fom the cente of a planet of mass M = 2.00 x 10 25 kg given the Univesal Gavitation law (F = Gm 1m 2 2 ). F = Gm 1m 2 2 F = 6. 67 x 10 11 x 1. 00 x 10 3 x 2. 00 x 10 25 (1. 65 x 10 7 ) 2 F = 4. 90 x 10 3 N (b) Show that the speed v of the space vehicle is given by the fomula v = Gm 2 F = Gm 1m 2 F = mv2 2 Gm 2 v 2 (Cancel out like tems) v = Gm 2 (c) Using the elationship fo the speed of the space vehicle given in pat (ii) and an expession fo the peiod T, showing that (T 2 = 4π2 3 Gm 2 ) v = 2π T v 2 = Gm 2 Squaing eveything in equation one yields: v 2 = 22 π 2 2 v 2 = Gm 2 T 2 Cancelling v 2 and eaanging the units of gives 4π 2 3 T 2 Gm 2 Reaanging this gives T 2 = 4π2 3 Gm 2
QUESTION 4 (2000 EXAM) (a) Calculate the magnitude of the gavitational foce F on mass m = 20.0 kg, positioned at 1.00 x 10 6 m above the Eath's suface. The mass of the Eath ME = 5.98 x 10 24 kg and the adius of the Eath RE = 6.38 x 10 6 m. = 1.00 x 10 6 + 6.38 x 10 6 = 7.38 x 10 6 F = Gm 1m 2 2 F = 6. 67 x 10 11 x 20 x 5. 98 x 10 24 (7. 38 x 10 6 ) 2 F = 146 N (b) A space vehicle of mass m is moving at a constant speed v in a cicula obit of adius ound the Eath. (i) Deive an expession fo v in tems of the adius and the mass of the Eath ME. F = GmM E F = mv2 2 GM E v 2 (Cancel out like tems) v = GM E (ii) Explain why the space vehicle does not need to use ocket engines to maintain its unifom cicula motion. Ignoe ai esistance. As the gavitation al foce is supplying the centipetal acceleation and this foce is constant.
QUESTION 5 (2001 EXAM) (a) State, in wods, Newton's law of univesal gavitation. Between any two masses, thee exists a mutual attactive foce. This foce is diectly popotional to the poduct of the masses and invesely popotional to the squae of the distance between them. This foce acts along a line between the centes of the masses. (b) Two masses exet a foce of magnitude F on each othe when placed a distance d apat. State the magnitude of the foce, in tems of F, if the distance d between the masses is doubled. F 1 2 if the distance wee doubled (x 2), the magnitude of F would be 4 times lowe in magnitude ( 1 its oiginal magnitude) 4 (1 mak) (c) Explain why the gavitational foces between two paticles of diffeent mass, M and m, as shown below, ae consistent with Newton's thid law. The foces ae equal in magnitude and opposite in diection. M exets a foce on m and hence, m exets the same amount of foce on M, just in the opposite diection. This is consistent with Newton s 3 d law
(c) Using Newton's second law and law of univesal gavitation, deive the expession (g = Gm 2 2 ) fo the gavitational acceleation at the Moon's suface. (M is the mass of the Moon, the adius of the Moon.) F = Gm 1m 2 2 F = mg Cancel out like tems (F and m) Gm 2 2 = g (d) Calculate the value of the acceleation due to gavity at the Moon's suface. (The mass of the Moon M = 7.35 x 10 22 kg and the adius of the Moon = 1.74 x 10 6 m.) g = Gm 2 2 g = 6. 67 x 10 11 x 7. 35 x 10 22 (1. 74 x 10 6 ) 2 g = 1. 62 ms 2 (e) The obit of a geostationay satellite ound the Eath is shown in the diagam below: (i) Explain why this satellite must obit in the same diection as the Eath otates. In ode to stay above the same location above the Eath, the satellite must obit in the same diection as the Eath is otating (1 mak)
(ii) Explain why the obit of this satellite must be equatoial. The cente of the satellites obit must be the cente of the eath as gavity is poviding the centipetal acceleation. Equatoial obits satisfy this condition and satellites in these obits otate with the same peiod as the Eath which allows the gavitational foce to act in the plane of the equato. (f) Explain why low-altitude pola obits ae used fo suveillance satellites. High esolution images ae available at lowe altitudes. The pola obits at lowe altitudes allow fo a geate numbe of successful evolutions of the Eath such that ove 24 hous, the satellite has photogaphed a lage % of the Eath s suface. Lowe obit allows fo less intefeence fom atmospheic gases and sunlight. (g) A space vehicle is moving at constant speed in a cicula obit A ound a planet, as shown in the diagam below: a v Daw and label vectos on the diagam to epesent the velocity v and acceleation a of the space vehicle at position P.
QUESTION 6 (2002 EXAM) (a) Calculate the magnitude of the gavitational acceleation g at the Eath's suface, using Newton's second law and the law of univesal gavitation. The mass of the Eath M is 5.98 x 10 24 kg and the adius of the Eath R is 6.38 x 10 6 m. g = Gm 2 2 g = 6. 67 x 10 11 x 5. 98 x 10 24 (6. 38 x 10 6 ) 2 g = 9. 80 ms 2 (b) A satellite obits the Eath with constant speed in a cicle whose adius is twice the adius of the Eath, as shown in the diagam below. The mass of the Eath is M= 5.98 x 10 24 kg and its mean adius is R = 6.38 x 10 6 m. (i) Show that the speed at which the satellite is moving is appoximately 6 x 10 3 ms -1. v = Gm 2 = 6.67x 10 11 x 5.98 x 10 24 (2 x 6.38 x 10 6 ) v = 5774. 9 ms 1 Rounding up to 1 sig. fig gives = 6 x 10 3 ms -1
(ii) Calculate the peiod of the satellite. Unlike planets which have elliptical obits, human-made satellites have cicula obits and theefoe cicula motion fomulae may be used: T = 2π v T = 2π(2 x 6.38 x 106 ) 5774.9 = 6.94 x 10 3 seconds (iii) Explain why the satellite tavels with unifom cicula motion in a fixed obit. The gavitational foce between the centes of the satellite and the lage mass is supplying the centipetal acceleation of the satellite. This is the ONLY foce acting on the satellite and theefoe accoding to Newton s 1 st law, whee the foces on a body ae constant, the speed of the body emains constant. A body moving with a constant speed in a cicula path is descibed as having unifom cicula motion.
QUESTION 7 (2003 EXAM) (a) Show that the speed v of a satellite moving in an obit of adius ound a planet of mass M is given by, (v = GM ), whee G is the gavitational constant. F = GmM mv2 2 F = GM v 2 (Cancel out like tems) v = GM (b) Explain the advantage of launching a low-altitude equatoial-obit satellite in a west-to-east diection. This type of launch takes advantage of the Eath s pe-existing otational velocity about its axis which occus in a west to eastely diection. The satellite will not need to gain any additional velocity to have an equatoial obit if launched fom west to east. This saves money which would be wasted on ocket fuel which would have to be used to gain the additional velocity (acceleate) should you be foolish enough to launch east to west.
QUESTION 8 (2004 EXAM) Two satellites, A and B, obit the Eath, as shown in the diagam below. Both satellites ae in cicula obits. The adius of satellite B is geate than the adius of satellite A. (a) On the diagam above, daw and label vectos to epesent the acceleation of satellite A and satellite B. (b) Satellite A obits at a adius of 2.112 x 10 7 m. Satellite B obits at a adius of 4.224 x 10 7 m and at a speed of 3072 ms -1. (i) Show that the speed v of a satellite moving in an obit of adius ound a planet of mass M is given by, (v = GM ), whee G is the gavitational constant. F = GmM mv2 2 F = GM v 2 (Cancel out like tems) v = GM
(ii) Hence show that the mass of the Eath is appoximately 5.98 x 10 24 kg. v = GM eaanged is M = v2 G M = (3072)2 x 4. 224 x 10 7 6. 67 x 10 11 M = 5. 976 x 10 24 kg theefoe, M = 5. 98 x 10 24 kg (iii) Calculate the obital speed of satellite A. v = GM v = 6. 67 x 10 11 x 5. 976 x 10 24 2. 112 x 10 7 v = 4. 334 x 10 3 ms 1 (iv) Calculate the obital peiod of satellite B. Unlike planets which have elliptical obits, human-made satellites have cicula obits and theefoe cicula motion fomulae may be used: T = 2π v T = 2π(4.224 x 107 ) 3072 = 8.639x 10 4 seconds (v) Geostationay satellites move in an equatoial obit in the same diection as the Eath's otation. Explain why geostationay satellites have obits of elatively lage adius. Geostationay obits emain above the same point on the Equato and so must have a peiod of evolution of 24 hous. The adius of the obit must be chosen such that the equied centipetal foce is exactly supplied by gavity, which esults in a elatively lage obit (35,900 km).
(vi) Two satellites of equal mass obit the Eath. One satellite has a adius of x and the othe has a adius of 2x (double the adius of x). Calculate the atio Fx : F2x of the gavitational foces acting on the satellites. Doubling the adius of the obit esults in 1 the gavitational foce due to 4 the invese squae law. F1 : F2 = 1 : 4 QUESTION 9 (2005 EXAM) Astonaut A is on the suface of a moon of adius. Astonaut B is at a distance of 3 fom the cente of the moon, as shown in the diagam below: Astonaut A and astonaut B have identical masses. The magnitude of the gavitational foce between the moon and astonaut A is 195 N. (a) Calculate, using popotionality, the magnitude of the gavitational foce between the moon and astonaut B. F = 1 2 F B = F A 3 2 F B = 195 9 F B = 21. 7 N
(b) A satellite is in a cicula pola obit aound the Eath at an altitude of 8.54 x 10 5 m. The mass of the Eath is 5.97 x 10 24 kg and its mean adius is 6.38 x 10 6 m. (i) Calculate the obital speed of the satellite. = adius of eath + altitude above Eath s suface = 8.54 x 10 5 + 6.38 x 10 6 = 7.234 x10 6 m v = GM v = 6. 67 x 10 11 x 5. 97 x 10 24 7. 234 x 10 6 v = 7. 42 x 10 3 ms 1 (ii) State two easons why low-altitude pola obits ae used in meteoology and suveillance. ANY TWO OF: They have highe esolution o poduce cleae images. They give almost global coveage. They have shot peiods so they can see pats of the Eath moe than once a day.
QUESTION 10 (2006 EXAM) Some satellites have geostationay obits and some satellites have pola obits (a) State two diffeences between geostationay obits and pola obits. A geostationay satellite stays above the same spot on eath wheeas a pola one does not A geostationay satellite obits above the equato, wheeas the pola one goes appoximately ove the poles OR: A geostationay satellite has a lage adius than a pola one. A geostationay satellite has a longe peiod than a pola one. (b) Deive the fomula (T 2 = 4π2 3 ) given the fomulas v = 2π and Gm 2 T v2 = Gm 2 v = 2π T v 2 = Gm 2 v 2 = 22 π 2 2 v 2 = Gm 2 T 2 4π 2 3 T 2 Gm 2 Reaanging this gives T 2 = 4π2 3 Gm 2 (c) Reaange this fomula to show that the peiod of satellite motion is given by the fomula 3 = Gm 2T 2 4π 2 T 2 = 4π2 3 Gm 2 3 = Gm 2T 2 4π 2 3 = Gm 2T 2 4π 2
(d) Hence detemine the altitude of a satellite in a geostationay obit aound the Eath. The mass of the Eath is M = 5.97 x 10 24 kg and its adius is R = 6.4 x 10 6 m. T = 24 x 3600 = 8. 64 x 10 4 s 3 = Gm 2T 2 4π 2 3 = 6. 67 x 10 11 x 5. 97 x 10 24 x (8. 64 x 10 4 ) 2 4π 2 3 = 6. 67 x 10 11 x 5. 97 x 10 24 x (8. 64 x 10 4 ) 2 4π 2 = 4. 227 x 10 7 m Altitude above Eath = 4. 227 x 10 7-6.4 x 10 6 = 3.6 x 10 7 m (4 maks) QUESTION 11 (2007 EXAM) The binay sta system known as Siius is shown, at one point in time, in the diagam below. The mass of Siius A, measued using data obtained fom the Hubble Space Telescope in 2005, is much lage than that of its patne sta, Siius B. (a) On the diagam above, daw vectos to show the gavitational foce acting on each of these stas at this point in time. Equal in length (by eye); Opposite in diection (towads one anothe) (b) Explain why Newton's law of univesal gavitation is consistent with Newton's thid law of motion. Newton s 3 d Law states that if Siius A exets a foce on Siius B, then Siius B will exet an equal but oppositely diected foce Siius B. The gavitational foce between two masses is attactive the foces ae oppositely diected and the magnitude of gavitational foce as pescibed by Newton's law of Univesal Gavitation, is diectly popotional to the poduct of the objects masses. The two foces will be the equal magnitudes as: M A x M B = M B x M A; Two foces act on diffeent objects.
(c) The pola-obiting satellite NOAA-N was launched in May 2005, as shown in the photogaph below: The satellite is now moving in a cicula obit above the Eath's suface at an altitude of 870 km. The mass of the Eath is 5.97 x 10 24 kg and its mean adius is 6.38 x 10 6 m. (i) Show that the obital speed of the satellite is 7.41 x10 3 ms -1 = 6.38 x 10 6 + 870 x 10 3 = 7.25 x 10 6 m v = GM = 6.67 x 10 11 x 5.97 x 10 24 7.25 x 10 6 v = 7. 411 x 10 3 ms 1 (ii) Calculate the magnitude of the acceleation due to gavity at the satellite's altitude. g = Gm 2 2 g = 6. 67 x 10 11 x 5. 97 x 10 24 (7. 25 x 10 6 ) 2 g = 7. 58 ms 2 (iii) Explain why the cente of the cicula obit of any Eath satellite must coincide with the cente of the Eath. The gavitational pull of the eath is towads the cente of the Eath. This is the net foce acting on the satellite and so this foce must cause centipetal acceleation. Theefoe must point towads the cente of the cicula obit of the satellite.
QUESTION 12 (2008 EXAM) A satellite of mass m moves in a cicula obit of adius about a planet of mass M, as shown in the diagam below. (a) Using Newton's law of univesal gavitation and Newton's second law of motion, show that the peiod T of the satellite is given by (T 2 = 4π2 3 Gm 2 ) F = GmM mv2 2 F = GM v 2 (Cancel out like tems) v = GM v = 2π T v 2 = Gm 2 v 2 = 22 π 2 2 v 2 = Gm 2 T 2 4π 2 3 Gm 2 T 2 Reaanging this gives T 2 = 4π2 3 Gm 2 (5 maks) (b) Two identical satellites X and Y, with obital adii 16 and espectively, move in cicula obits about the Eath, as shown in the diagam below. (i) Using popotionality, calculate the atio Tx : Ty of the satellites' obital peiods. If, T 2 = 4π2 3 Gm 2 then T 2 3 T 3 Tx : Ty = x 3 y 3 Tx : Ty = 16 3 y 3 Tx : Ty = 4096 1 Tx : Ty = 64 : 1
(ii) Satellite X is a geostationay satellite moving in the Eath's equatoial plane. Explain why this satellite must move in a paticula obit of elatively lage adius. Since T 2 3 o since T depends on As T = 24 hous which is a elatively lage peiod, then R has to be elatively lage. QUESTION 13 (2009 EXAM) The centipetal acceleation of the Eath in its obit aound the Sun has a magnitude of 5.90 10 3 ms 2. This acceleation is caused by the gavitational foce that the Sun exets on the Eath. The mass of the Eath is 5.97 10 24 kg and the mean adius of the Eath s obit aound the Sun is 1.50 10 11 m. (a) Show that the magnitude of the gavitational foce that the Sun exets on the Eath is 3.52 10 22 N. F = ma F = 5. 97 x 10 24 x 5. 90 x 10 3 F = 3. 52 x 10 22 N (b) Hence detemine the mass of the Sun. F = Gm 1m 2 2 m 2 = F2 Gm 1 m 2 = 3. 52 x 1022 x (1. 50 x 10 11 ) 2 6. 67 x 10 11 x 5. 69x 10 26 m 2 = 1. 99 x 10 30 N
(c) Two satellites ae moving in cicula obits aound the Eath, Obit A and Obit B, as shown in the diagam below. The adius of Obit B is double the adius of Obit A. Speed of satellite in obit A (i) Calculate the atio Speed of satellite in obit B v = GM v 1 v a v b = 1 a 1 b (since G and M ae the same fo both satellites you can cancel them out) a b = a 2 b (cancel out fom both) = 2 (ii) Explain why the centipetal acceleation of a satellite in an obit of constant adius is independent of the mass of the satellite. v = GM Since the speed of the satellite depends only on the mass of the eath and not the mass of the satellite, then the centipetal acceleation is also independent of the mass of the satellite.
(d) The diagam below shows two cicula paths aound the Eath. Path 1 is not a possible satellite obit. Path 2 is a low-altitude pola obit. (i) Explain why Path 1 is not a possible satellite obit. The cente of the obit in the case of path 1 does not coincide with the cente of the eath. The satellite's obit cannot be stable. This is because the gavitational foce must act towads the cente of the eath and towad the cente of the obit in ode to povide the centipetal acceleation fo unifom cicula motion. (ii) State one eason why satellites in low-altitude pola obits ae often used fo suveillance, and explain you answe. Low-altitude obits ae close to the suface of the eath. This poduces geate esolution. This means that images show geate detail and ae theefoe moe useful fo suveillance.
(e) The Iidium satellite netwok consists of sixty-six communication satellites obiting the Eath at a adius of 7.18 10 6 m, with a speed of 7.46 10 3 ms 1. On 10 Febuay 2009 the Iidium 33 satellite collided with the Russian Cosmos 2251 satellite above Sibeia. The damaged Iidium 33 satellite was eplaced in the netwok by a spae satellite that was obiting at a lowe adius. Calculate the time that a satellite in the Iidium netwok takes to complete one obit of the Eath. Give you answe to the coect numbe of significant figues. T = 2π V (since obit is appoximately cicula as it is a human-made satellite) T = 2π7. 18 x 106 7. 46 x 10 3 T = 6.05 x 10 3 sec QUESTION 14 (2010 EXAM) (a) Explain why a geostationay satellite must obit in a west-to-east diection. A geostationay satellite is one that emains ove a fixed point of the eath's suface. This can only occu if it tavels in the same diection that the eath otates (ie west-to-east). (b) Explain one advantage of launching equatoial-obit satellites in a west-to-east diection. By launching a satellite in the same diection that the eath spins, the speed of the eath's spin contibutes to the launch speed of the satellite. This makes the launch moe economical as it saves on enegy and hence fuel.
The diagam below shows two isolated, spheically symmetic objects. The mass of Object A is much lage than the mass of Object B. On the diagam above, daw vectos to show the gavitational foces that these objects exet on each othe. HINT: Despite one mass being lage, the gavitational foce acting on each object is equal in magnitude but opposite in diection. QUESTION 15 (2012 EXAM) In Novembe 2012 pats of the wold will expeience a total sola eclipse. Duing such an eclipse the Eath, the Moon, and the Sun ae in a staight line. The Moon is between the Eath and the Sun. In this alignment the distance between the Eath and the Moon is 3.85 x 10 8 m, and the distance between the Moon and the Sun is 1.50 x 10 11 m. The mass of the Eath is 5.97 x 10 24 kg. The mass of the Moon is 7.35 x 10 22 kg. The mass of the Sun is 1.99 x 10 30 kg. foce on the moon due to eath Detemine the magnitude of the atio foce on moon due to sun foce on the moon due to eath F = Gm 1m 2 2 F = 6. 67 x 10 11 x 7. 35 x 10 22 x 5. 97 x 10 24 (3. 85 x 10 8 ) 2 F = 2. 00 x 10 20 N foce on the moon due to sun F = Gm 1m 2 2 F = 6. 67 x 10 11 x 7. 35 x 10 22 x 1. 99 x 10 30 (1. 50 x 10 11 ) 2 F = 4. 00 x 10 20 N (4 maks) foce on the moon due to eath foce on moon due to sun = 2.00 x 1020 N = 0.5 So atio is 1 : 2 4.00 x 10 20 N
The Quick Bid satellite is used to ceate images of the Eath. One such image is shown below left. The satellite obits at an altitude of 482 km, and has a mass of 9.5 x 10 2 kg. The Intenational Space Station (shown in the image below ight) obits at an altitude of 390 km, and has a mass of 4.2 x 10 5 kg. (a) State whethe the QuickBid satellite obits the Eath at a faste o slowe speed than the Intenational Space Station. Give a eason fo you answe. Quickbid obits FASTER. v = GM Since the adius of quickbid is much smalle in magnitude (being close to the cente of the Eath), the magnitude of velocity will have to be lage to maintain an obit as the adius is invesely popotional to the velocity of the satellite. (b) State any effect that the diffeent masses of the satellites will have on thei speeds. Give a eason fo you answe. v = GM Since the speed of the satellite depends only on the mass of the eath and not the mass of the satellite, then the centipetal acceleation is also independent of the mass of the satellite. (c) State one advantage of the QuickBid satellite's low-altitude obit. Low-altitude obits ae close to the suface of the eath. This poduces geate esolution. (1 mak)