using analyse planetay and satellite motion modelled as unifom cicula motion in a univesal gavitation field, a = v = 4π and g = T GM1 GM and F = 1M SATELLITES IN OBIT A satellite is any object that is in a stable obit aound anothe object. The Eath and all the othe planets ae natual satellites of the Sun. The moon is the Eath's only natual satellite but thee ae thousands of atificial satellites obiting. These ae used fo communication, weathe foecasting, geological suveying and espionage. Fo a satellite in a stable cicula obit, the only foce acting on a satellite is the gavitational attaction between it and the cental body. This foce acting on it is always pependicula to its motion. Theefoe the enegy of the satellite is unchanged as it obits. The kinetic enegy and gavitational potential enegy both stay the same. The foce of gavity holds the satellites in thei obits and causes them to have an acceleation towads the cental mass. If the foce of gavity could be 'tuned off' then the satellites would fly off at a tangent. The satellites ae continually falling towads the eath at the same ate that the cuved suface of the eath is falling away fom the satellites. Since satellites ae in a continual state of fee-fall, thei acceleation will equal the gavitational field stength at that point. v a = 4 GMm GM Using v = a = Using F = and F = mg T T = g whee v = speed (m/s) = adius of obit (m) T = peiod of obit (s) M = cental mass (kg) g = gavitational field stength (N/kg). Notice that the acceleation of the satellite is independent of the mass of the satellite. If a satellite is in a geo-stationay obit, the satellite obits the Eath evey 4 hous. This is a 'synchonous' obit and keeps the satellite above the same place all the time. Yea 1 Physics Page 1
- PAST VCAA EXAM Examples 00 mass of Cassini. 10 3 kg mass of Jupite 1.9 10 7 kg mass of Satun 5.7 10 kg Satun day 10.7 hous When Cassini aives in the vicinity of Satun this yea, scientists want it to emain above the same point on Satun s equato thoughout one complete Satun day. This is called a stationay obit. Question 5 What is the peiod in seconds of this stationay obit? Satellite needs to have same peiod as a day on Satun. ie 10.7 days. In seconds this equals 3.85 x 10 4 s. Examine s comment To emain above the same point on Satun s equato the satellite would be equied to have a peiod of 10.7 hous, o 3.85 x 10 4 s. The main difficulties wee to eithe assume a 4-hou day o to make an aithmetic eo in the calculation. Question Calculate the adius of this stationay obit. (G =.7 10-11 N m kg - ) You need to use the vaiation of the usual equation So the equation becomes On substitution this becomes GM 4π. T GMm m4π. In this case, the mass m of Cassini cancels out. T 11.7 10 5.7 10 4π (3.85 10 ) 3 = 4 11 4.7 10 5.7 10 (3.85 10 ) 3 = 1.44 10 4 = 1.13 10 8 m 4π Yea 1 Physics Page
Examine s comment Application of Newton s Law of Univesal Gavitation fo the foce between two masses along with the elation GMm m4π fo unifom ciculation motion esulted in the equation: T Substitution of the appopiate values esulted in a adius of 1.1 x 10 8 m fo the stationay obit. Many students expeienced difficulty with the concept of a stationay obit. Othes had difficulty getting stated, often stating with Newton s Law of Univesal Gavitation but wee unable to combine this with the cicula motion equation mv m4π involving the peiod. Students ae moe comfotable with the elation but not so familia with T Fo those who could successfully wite down and substitute into the fomula, many made aithmetic eos. The final stage of taking the cube oot to find was vey pooly done. 001 The Mi space station obited Eath at an altitude of 390 km. The total mass of the space station was 140 tonnes and Mi completed 83 500 obits in the 14. yeas befoe it cashed to Eath on 3 Mach 001. Question 1 Calculate the peiod of Mi s obit in seconds. If the space station completed 83 500 obits in 14. yeas, then he peiod is given by 14. 35 4 0 0 = 5.514 10 3 secs 5.5 10 3 s 83500 time(secs) obi ts = Examine s comment Ave (1.5/3) The peiod of Mi s obit was 5.5 x 10 3 m s -1. This was a elatively simple question and the most common poblems wee eithe due to simple aithmetic mistakes o to substituting of incoect numeical values when conveting 14. yeas into seconds. Some students took a ound-about appoach and went back to the equations fo Univesal Gavitation and cicula motion, often making aithmetic eos when doing so. Yea 1 Physics Page 3
Question Calculate the speed of Mi while in obit. You need to be caeful with the adius of obit. The station is 390 km above the suface of the Eath, so the adius of obit is E + 390 km. The units ae mixed, so you need to be caeful. obit =.37 10 m (.37 10 390 10 ) v = = = 7703 ms -1 = 7.7 10 3 ms -1 t 3 5.5 10 3 Examine s comment Ave: (1.5/3) The speed of Mi while in obit was 7.7 x 10 3 m s -1. This question was easonably well answeed, paticulaly by those students who woked with the given data athe than tying to wok with equations fom the data booklet o thei A4 sheet. Common eos esulted fom failing to include the altitude when detemining the adius of the obit and the usual poblem of calculating aithmetic expessions involving powes of 10. Question 3 Calculate the value of the gavitational field stength at 390 km above Eath s suface. Since it is in cicula motion, a = g = (7.7 10.37 10 3 ) 390 10 3 v whee g is the gavitational field stength. 59.9 10 g = 8.78 Nkg -1.7 10 Examine s comment Ave: (./5) Fo unifom cicula motion the acceleation can be calculated via application of the equation a = v /, using the speed calculated in Question and the adius of the obit as the adius of the eath plus the altitude. This esulted in a gavitational field value of 8.8 N kg -1. The question itself was easonably well undestood by most students. The two most common eos wee in failing to include the altitude when detemining the obit adius and using the mass of Mi athe than Eath fo the calculation. Students need to be awae that it is the cental mass, not that of the obiting body, that detemines gavitational field stength and they need to be eminded to think moe caefully when choosing values to substitute into fomulas. When Mi was in stable obit, an object of mass.5 kg was placed on a sping balance that was attached to the inside of the space station. Question 4 What would be the weight shown on the sping balance? Explain you easoning, o show you woking, fo detemining the eading of the sping balance. Since Mi is in a stable cicula obit, and a = g, then the eaction foce fom the sping balance should be zeo. The net foce = mg T. This must equal ma. But a = g, so T = 0. O mg T = ma mg = ma = T T = 0. Yea 1 Physics Page 4
Examine s comment Ave: (./5) This question was testing the concept of appaent weightlessness in a slightly diffeent fomat than in pevious examinations. Both the sping and the mass ae acceleating at the same ate and hence the eading of the sping balance will be zeo. This poved to a easonably demanding question and cetainly highlighted just how pooly students undestand the concept of appaent weightlessness. The most common eo was to give the answe as the weight foce mg ( N). Vey few students mentioned that the net foce on the mass was mg and hence the sping foce was zeo. Vey few chose to daw a foce diagam and use this to calculate the sping foce as zeo. Many answes efeed to feefall o appaent weightlessness but without using this to eally addess the question asked. Many students inappopiately used the tem nomal eaction athe than tension when efeing to the foce exeted by the sping. 000 When people went to the Moon in the Apollo 11, the spacecaft was initially placed in a paking obit 190 km above Eath s suface. This is shown below. Question 3 Calculate the speed of Apollo 11 in the paking obit. ( E =.37 10 m, M E = 5.98 10 4 kg, G =.7 10-11 N m kg - ) In ode to do this question you need to undestand that the net foce on the spacecaft is the gavitational foce, and that it is this gavitational foce is the centipetal foce equied to keep the spacecaft in obit. Theefoe you can wite; F net GMm v m divide both sides by m GM v times both sides by GM v take the squae oot of both sides GM v Befoe you sub the values in make sue they ae the ight ones Lets establish what is; it is the distance fom the cente of mass of the Eath to the cente of mass of the spacecaft. So it s the adius of the eath plus the distance the spacecaft is above the Eath s suface. 5 5 E 1.9 10.37 10 1.9 10.5 10 m M is the mass of the eath = 5.98 10 kg Now you can sub the values in and you get v.7 10 11 5.98 10.5 10 4 4 v 7.8 10 3 ms 1 Yea 1 Physics Page 5
s The paking obit of Apollo 11 is an example of unifom cicula motion in which the net foce is povided by the gavitational foce between Eath and Apollo 11. This esulted in a speed of 7.8 10 3 ms -1. The aveage mak of 1.7/3 demonstates that just ove 50% students had a easonable gasp of this concept. In fact, about half of the students scoed the full 3 maks fo this question, but about 30% scoed zeo. The most common eo was made by students fogetting to add the spacecaft s altitude to the adius of Eath when detemining the adius of the obit. Many students made simple eos when substituting in numeical values o when using thei calculato. Two students wee discussing the physics of the motion of Apollo 11 in the paking obit. They ealised that it was tavelling at a constant speed but not in a staight line. Jane said that this can be pedicted by Newton s fist law. Maia claimed that this disobeyed Newton s fist law. Question 4 Explain, giving easons, whethe Jane is coect o incoect, and whethe Maia is coect o incoect. A body tavelling in a staight line will continue to tavel in a staight line unless a net foce acts on it. the velocity of Apollo 11 was changing (change in diection), it was acceleating if Apollo 11 was acceleating thee must have been a net foce acting on Apollo 11. Jane was coect, and Newton s fist Law was not boken. You must answe the question by making statement about Jane and/o Maia In ode to addess whethe Jane o Maia was coect students wee expected to discuss the following points: a statement of Newton s Fist Law the velocity of Apollo 11 was changing (change in diection), hence it was acceleating a net foce (gavitational foce) was acting on Apollo 11. The aveage scoe hee was 1.1/3. It is clea that students ae able to ecall Newton s Laws but wee unable to apply them in a paticula physical situation. Few students discussed the net foce acting on Apollo 11 o the fact that it was acceleating. Question 5 Which one of the following statements (A D) best descibes the oigin of the centipetal foce equied to keep Apollo 11 in its cicula paking obit? A. Apollo 11 s acceleation towads Eath B. the ocket motos of Apollo 11 C. the speed of Apollo 11 in its obit D. the gavitational field of Eath acting on Apollo 11 This question is asking whee is the centipetal foce equied to keep the spacecaft in obit coming fom. The only statement that answes this question is D, the gavitational field of the Eath acting on Apollo 11. The oigin of the net foce needed to suppot unifom cicula motion of Apollo is that of the gavitational attaction exeted by Eath on Apollo 11. Hence, statement D was coect. The aveage mak of 0.57/1 was a disappointing esult fo such a simple question. Yea 1 Physics Page
Apollo 11 then leaves its paking obit and tavels to the Moon. Question Which one of the following gaphs (A F) best epesents the net gavitational foce acting on Apollo 11 as it tavels fom its paking obit to the Moon? Gaph C best epesents the net foce acting on Apollo 11 as it tavels fom the paking obit to the moon. The things to know would be the foce is not linea, so B, E and F ae out, when the foce fom the Eath equals the foce fom the Moon, the spacecaft will be close to the moon then the Eath, and it will not occu half way between the two objects. This ules out A and D theefoe C is the most easonable answe. Gaph C best epesents the vaiation of the net gavitational foce acting on Apollo 11. The foce deceases accoding to the invese squae law as Apollo 11 tavels away fom Eath, eaching zeo at a point close to the Moon. The diection of the foce now changes as Apollo 11 expeiences a net foce diected towads the moon. The magnitude of this net foce again inceases as Apollo 11 appoaches the Moon. The aveage mak hee was 1.17/, which was easonable fo what was anticipated to be a moe difficult question. Yea 1 Physics Page 7
1999 Nato III is a communication satellite that has a mass of 310 kg and obits Eath at a constant speed at a adius of 4. 10 7 m fom the cente of Eath. Question 1 Calculate the magnitude of Eath s gavitational field at the obit adius of Nato III. Give you answe to thee significant figues. You must show you woking. The gavitational field stength is given by Substituting gives g = g = 0.4 ms - GM g e.7 10-11 7 (4. 10 5.98 10 ) 4 = 0.397 The magnitude of Eath s gavitational field calculates out to 0.4 N kg -1. This equied a simple substitution into the field stength equation and with an aveage scoe of 1.5/ a majoity of the students wee able to do this. The vast majoity of students ecognised the coect fomula to use and any eos wee due to using the incoect value fo the mass o being unable to coectly manipulate the data when using the calculato. A numbe of candidates failed to answe to thee significant figues as equested in the question. Question What is the speed of Nato III in its obit? The gavitational field stength equals the acceleation of the satellite. Thus v 3 1 0.4 v 3.07 10 ms The speed of Nato III calculates to 3.1 x 10 3 m s -1. This could have been achieved by simply combining the fomulas of Univesal Gavitation and unifom cicula motion. The aveage mak fo this question was 1.9/3 indicating that the majoity of candidates undestood how to deive the necessay equation and then coectly calculate the speed. The equation fo the speed can be deived diectly and many students did so. Howeve, it was disappointing to note the numbe of students who stated by calculating the peiod of the obit and then detemining the speed in what was a two-step pocess. This latte method not only wasted valuable exam time but inceased the likelihood of an aithmetic eo as well. This question also highlighted the continuing poblem that many students have when using the calculato to detemine the value of expessions with indices. Yea 1 Physics Page 8
Question 3 Which one of the following statements (A D) about Nato III is coect? A. The net foce acting on Nato III is zeo and theefoe it does not acceleate. B. The speed is constant and theefoe the net foce acting on Nato III is zeo. C. Thee is a net foce acting on Nato III and theefoe it is acceleating. D. Thee is a net foce acting on Nato III, but it has zeo acceleation. C If thee was no net foce on the satellite it would tavel in a staight line athe than obit. Nato III is tavelling in unifom cicula motion and hence thee is a net foce, povided by the gavitational attaction to Eath, acting towads the cente of the obit (Eath). Thus, statement C was coect. The aveage mak of 1.0/ demonstates that only 50% of candidates coectly undestood this concept. Yea 1 Physics Page 9