Gaia s Place in Space

Transcription

1 Gaia s Place in Space The impotance of obital positions fo satellites Obits and Lagange Points Satellites can be launched into a numbe of diffeent obits depending on thei objectives and what they ae obseving. Many satellites ae launched into an obit in which they cicle aound the as it obits the (the Moon also does this!). Howeve, this is not the only viable obit fo a satellite. Stable obits ae also possible at special locations known as Lagange points. At these obital locations, satellites do not cicle the but athe obit the in sync with it. The five Lagange points, named afte the astonome and mathematician, Joseph-Louis Lagange, ae positioned at locations which ae made stable fo satellite obsevations due to a balance of gavitational foces and obital motions of the and the. Let s take a look at each of the points individually. Lagange point 1 (L 1 ) as shown in Figue 1 follows an obital path that is close to the than s obit. Figue 1 The obital path of a satellite located on a Lagange point 1 obit Satellite at L 1 Step 1: Show how the Newton (N) is the unit of measuement fo the gavitational foce that is calculated between two objects in the following equation: F g = foce due to gavity G = gavitational constant m 1 = mass of cental body m 2 = mass of obiting body = distance between the two bodies F g = G m 1 m 2 2

2 F g = G m 1 m 2 2 Which cancels into: F = kg m s 2 = N = m3 kg 1 s 2 x kg x kg m 2 Step 2: What gavitational influences would you expect an object located in the L 1 position to expeience? Students should ecognise that this object would expeience gavitational influences fom both the and. Step 3: What is the definition of a centipetal foce? A centipetal foce is a foce that causes an object to follow a cicula motion, pulling the object towads the cente of this cicle. This is demonstated in the equation: F c = m v2 F c = centipetal foce (N) m = mass of obiting body (kg) v = obital velocity (m s -1 ) = distance between the two bodies (m) Step 4: Using the equation povided and you knowledge of Newton s Law of Gavitation, fo an object located close to the, how would you expect its obital velocity to diffe fom s? F c = centipetal foce (N) m = mass of obiting body (kg) v = obital velocity (m s -1 ) = distance between the two bodies (m) F c = m v2 Students should ecognise that they need an equation which elates obital velocity (v) to distance () and which may include othe known o constant quantities. Then they will be able to see how v vaies with. In the fomula povided, F c is unknown so we need to eplace it by something which is known. Students should then ecognise that the centipetal foce holding a satellite o planet in obit aound the is the gavitational foce (F g ). Thus, F c can be eplaced in the fomula povided by the fomula fo the

3 gavitational foce. This gives an equation fo how v elates to with all othe quantities being known o constant. F g = Gm 1m 2 2 F g = foce due to gavity (N) G = gavitational constant (6.67 x m 3 kg -1 s -2 ) m 1 = mass of cental body (the ) (kg) m 2 = mass of obiting body (kg) = distance between the two bodies (m) Replacing F c with the gavitational foce we obtain: F g = Gm 1m 2 2 = m 2 v2 If we then eaange the equation to make obital velocity (v) the subject: v = Gm 1 We can now see that as you get close to the and deceases, the obital velocity (v) of the object will incease. Howeve, an object on an L 1 obit does not ovetake the on its obit. L 1 is positioned at a specific distance of 1.5 x 10 6 km fom the so that the exta influence of gavity that the object should expeience fom the is cancelled out by the gavitational pull fom in the opposite diection. As a esult, the object s obital speed is educed and theefoe it stays in pace with the. Like L 1, Lagange point 2 (L 2 ) is also located 1.5 x 10 6 km away fom. Howeve, L 2 lies futhe fom the, placing L 2 outside s obit athe than inside like L 1. This is illustated in Figue 2. Figue 2 A satellite in an obit at the second Lagange point, L 2. Satellite at L 2

4 Step 5: Fo an object located futhe away fom the, how would you expect its obital velocity to diffe fom s? Students should ecognise that when compaing obits of diffeent distances, you would expect the object that lies at a distance futhe fom the to have a slowe obital velocity and theefoe the close object would ovetake the object that lies futhe away. This is due to the weake influence of the s gavity at geate distances in accodance with the equations detailed in Step 4. Howeve, again at this specific distance of 1.5 x 10 6 km, the gavitational influence exeted by pulls the object aound its obit, speeding it up and allowing it to keep pace with the obit of. Lagange point 3 (L 3 ) is located at much geate distance fom, positioned behind the, diectly opposite s obital position. As shown in Figue 3, a satellite located hee would follow the same obit as the just on the opposite side of the. Figue 3 The obital path of a satellite at Lagange point 3. Satellite at L 3 Step 6: Fo an object positioned at the L 3 point, how would you expect its obital velocity to compae to that of? Again due to the equation fomulated in Step 4: v = Gm 1 v = obital velocity (m s -1 ) G = gavitational constant (6.67 x m 3 kg -1 s -2 ) m 1 = mass of cental body (the ) (kg) = distance between the two bodies (m) We see that the obital velocity vaies only with distance () fom the. (In fact, the and othe massive Sola System bodies will also have some gavitational influence on the satellite but this will be extemely small compaed to the gavitational influence of the, which is by fa the dominant effect hee).

5 Students should ecognise that the object lies at the same distance fom the as the and theefoe has the same obital velocity as the. The object will continue to obit the at the same pace as the but on the opposite side of the. Lagange points 4 and 5 (L 4 and L 5 ) ae positioned at 60 o ahead (L 4 ) and behind (L 5 ) of s obit, see Figue 4. Figue 4 The Lagange points 4 and 5 in elation to s obital position. Satellite Satellite 2 Step 7: The is obiting the in an anti-clockwise diection in Figues 1-4. Which Lagange points do satellites 1 and 2 coespond to in Figue 4? Satellite 1 is positioned at Lagange point 4 and Satellite 2 is positioned at Lagange point 5. To take a look at some animations of obits at the Lagange points, follow the link below: Figue 5 summaises all five of the Lagange points. Figue 5 The five Lagange point obits. L 4 L L 1 L 5 L 3

6 Now you ae familia with the diffeent obital positions known as the Lagange points, we can apply them moe specifically to Gaia. Step 8: Refe to the Intoducing Gaia woksheet. Given what you know about the Gaia satellite and what its mission objectives ae, which of the Lagange points do you think would be the most suitable fo Gaia s position? Justify you answe. Students should ecognise that L 2 would be the best location fo Gaia. This is because the light fom the, and Moon on one side of Gaia ae blocked by the satellite s sunshield, giving Gaia an unobstucted view of the sky in the othe diection. With these souces of backgound light blocked fom getting into its instuments, Gaia is moe sensitive to light fom fainte distant souces and is capable of making moe accuate measuements. This position also means Gaia is moe easily able to stay cool as it expeiences lowe levels of heat fom the than it would on an L 1 obit. L 1 and L 2 ae positioned close to the than the othe Lagange points. This stands in favou of data tansmission ates. Step 9: You might expect Gaia to fall into the s shadow due to s position between Gaia and the. Watch the video below to see how Gaia avoids enteing the s shadow: Hee, Gaia not only keeps pace with s obit aound the, it also follows an obit aound the L 2 point evey 180 days. This means that Gaia can avoid being eclipsed by and going into themal shock as a esult of sudden changes in tempeatue. This is also illustated in Figue 6. Figue 6 Gaia s obit aound L 2. This diagam is NOT to scale. Shadow 45 o Gaia at L 2 Gaia s Obit aound L 2 Step 10: The Moon is appoximately 382,500 km away fom. How many times futhe away fom is Gaia at the L 2 point? 1.5 x 10 6 km 382,500 km 382,500 km = 3.9 L 2 is 3.9 times futhe away fom than the Moon.

7 Step 11: How many km futhe away fom us is Gaia than the Moon? 1.5 x 10 6 km 382,500 km = 1,100,000 km L 2 is ove 1 million km futhe away fom than the Moon is. Step 12: The Moon itself is appoximately 3,476 km in diamete. How many Moons could fit between the and Gaia at L 2? 1.5 x 10 6 km 3,476 km = 430 You could fit appoximately 430 Moons in the space between the and Gaia. Step 13: Use the infomation povided to calculate the foce of gavitational attaction between Gaia and the. Gavitational Constant = 6.67 x m 3 kg -1 s -2 Gaia Distance = x 10 8 km Mass of = 1.99 x kg Mass of Gaia = 2,030 kg Students should again begin by conveting the distance value fom km to m. This is because foce is measued in units of kg m s -2 : x 10 8 km = x m Then by inputting the coesponding values into Newton s Law of Gavitation we obtain: F g = G x 1.99 x 1030 kg x 2,030 kg (1.511 x m) 2 = 11.8 N The foce of gavitational attaction between the and Gaia is appoximately 12 N. Step 14: Ignoing the additional obit of Gaia aound L 2 and accounting only fo its obital path aound the, what is the total distance coveed by Gaia in one obit? (State you answe to 4 significant figues). s Distance fom the = x 10 8 km Students should ecognise that hee the total distance is calculated using the equation fo the cicumfeence of a cicle: cicumfeence = 2 x π x adius As Gaia is positioned at L 2, outside the obit of, they must ecognise to add Gaia s distance fom to the s distance fom the :

8 1.496 x 10 8 km x 10 6 km = x 10 8 km Theefoe by inputting the value calculated above into the equation fo calculating the cicumfeence, we obtain: obital distance = 2 x π x 1.511x 10 8 km = x 10 8 km Step 15: Gaia obits the in appoximately the same time as the does. Use you answe fom Step 14 to calculate the speed of Gaia s obit aound the. Give you answe in km s -1. Students should ecognise that they will need to apply the following equation: speed = distance time They should fist ecognise that the takes appoximately 365 days to obit the and convet this into units of seconds: 365 days x (24 x 60 x 60 s pe day) = 31,536,000 s Then by inputting the coesponding values into the above equation we obtain: speed = x 108 km 31,536,000 s = km s 1 Gaia obits the at a speed of aound 30 km s -1, which is appoximately the same as s. Step 16: Use you answe in Step 15 and the equation below to calculate the centipetal foce on Gaia s obit aound the. F c = centipetal foce (N) m = mass of obiting body (kg) v = obital velocity (m s -1 ) = distance between the two bodies (m) F c = m v2 Students calculated Gaia s velocity in Step 15 to be km s -1. Theefoe by inputting the othe infomation that is povided we obtain: 2,030 kg (3.011 x 104 m s 1 ) x m = 12.2 N The centipetal foce acting on Gaia s obit aound the is appoximately 12 N.

9 Step 17: How does the centipetal foce you calculated in Step 16 compae with the value you obtained fo the foce of gavitational attaction in Step 13? Students should have calculated answes which ae close to each othe, but with the value of the centipetal foce calculated in Step 16 being highe. This is because the gavitational foce on Gaia also contains contibution fom the, which was not included in the calculation in Step 13. Even though the is much less massive than the, because it is elatively close to L 2 its gavitational influence will count hee in fact, as we have seen, the s influence is what allows a satellite placed at L 2 to keep pace with the even though it is on a wide obit. In this case, the centipetal foce on Gaia is povided by a combination of the gavitational foces fom the and fom the. Step 18: Assume the aveage unning speed of a human is 4.0 m s -1. How does this compae with the velocity of Gaia s obit aound the? Students should have calculated the speed of Gaia s obit aound the to be 30 km s km s 1 = 30,000 m s 1 Theefoe: 4 m s 1 ( 30,000 m s 1) x 100 = 0.01 % Humans can un at a speed equal to 0.01 % of Gaia s obital velocity Step 19: If you wee to un the distance of Gaia s obit aound the at 4.0 m s -1, how long would it take you to complete one obit? Give you answe in yeas. Hint: You will need to use you answe fom Step 14. Fom Step 14, students should have calculated the distance of Gaia s obit aound the to be x 10 8 km x 10 8 km = x m The time it would take fo a human to un this distance is calculated using the following equation: speed = distance time Theefoe: x m 4.0 m s 1 = 2.4 x s To convet this into yeas we know that:

10 1 y = (60 x 60 x 24 x 365) s Theefoe: 2.4 x s (60 x 60 x 24 x 365) s pe y = 7,600 ys Step 20: How does you answe fom Step 19 compae with Gaia s mission lifetime and a human lifetime? It would take a human appoximately 7,600 yeas to complete just one of Gaia s obits. Theefoe, Gaia would have long finished its mission lifetime, not to mention this is fa longe than any human lifetime. Step 21: What consideations need to be made when deciding on an obit fo a satellite? Answes hee ae at the teache s discetion, howeve some of the main consideations ae: What it is you ae obseving how bight o how faint it is. Is it impotant fo the satellite to be kept cool? Does the satellite need to be shielded fom the?