C Un satélite de telecomunicaciones de 5 000 kg de masa describe una órbita circular concéntrica con la Tierra a 1200 km de su superficie. Calcula: a. La velocidad orbital del satélite. b. Su período de revolución. Solución COMPRENSIÓN. m = 5 000 kg
h = 1,2 ⋅ 106 m
6 m de la Tierra más El radio es=igual radio r = R T + h = 6,37 ⋅ 106 mde + la 1,2órbita ⋅ 106 m 7,57 al ⋅ 10
la distancia del satélite a la superficie terrestre: 6 m + 1,2 ⋅ 6106 m = 7,57 ⋅ 6106 m r =+ R Th+=h6,37 = 6,37 r = RT ∙ 10⋅ 610 m+ 1,2 ∙ 10 m = 7,57 ∙ 10 m 6 m + 1,224⋅ 106 m = 7,57 ⋅ 106 m 2 r = R + h = 6,37 ⋅ 10 m = 5 000 5,98 ⋅ 10orbital kg del satélite: M Tkg a.T Calculamos m velocidad −11 N ⋅ la 6 6 v= G = 6 ,67 ⋅ 10 ⋅6
⋅ 10 m + 1,26 ⋅ 10 m = 7,57 ⋅ 10 m r r = R T + h = 6,37 m = 5 000 kg kg 2 6 7,57 ⋅ 10 m 6 6 6 r = R + h = 6,37 ⋅ 10 m + 1,2 N ⋅ 10 = 7,57 m R T = 6,37 ⋅ m h ⋅ 10 kg T ⋅ 10 ⋅2410 kg m m==1,2 5 000 M ⋅ m 2 m 5,98 −11 v v==7, 3G⋅ 10T3 m/s = 6 ,67 ⋅ 10 ⋅ 6 2 m = 5 000 kg r 7,57 ⋅ 1024 2 6 m 6 m + 1,2 ⋅kg 6 m h = 1,2 ⋅ 106 m r = R + h = 6,37 ⋅ 10 10 = 7,57 ⋅ 10⋅ 10mkg 5,98 M N ⋅ m T 6 −11 T R T =6 6,37 ⋅ 10 m 3 2 ⋅ 5,98 ⋅ 10 24 kg v = G = 6 , 67 ⋅ 10 m h = 1,2 ⋅ 10 v = 7−, 11 3 ⋅N 10⋅ m MT 2m/s 6 7,57 ⋅ 10 6 m6 m = 5 000 kg MT = 5,98 ⋅ 1024 kg 6 m + kg 6 24 6 m v π= G2M 6m ,67 ⋅ 10 RπrT⋅ 7,57 + h== 6,37 ⋅ 10 1,2 ⋅2210 ⋅ 5,98 m = 7,57 ⋅ 610 m h = 1,2 ⋅ 10 2 r r= ⋅10 ⋅ 10 kg 3 N ⋅ m 6 − 11 rT = el R T = 6,37 ⋅ 10 m ⋅ m 7,57 10 T =v = b.=G = 6 , 5 ⋅ 10 s kg Calculamos período de revolución: 6 , 67 ⋅ 10 ⋅ 3 m = 5 000 kg v v = 7, 3 ⋅ 10 kg m/s 2 2 6 m mT = R T = 6,37 ⋅ h10=61,2 r M⋅ 10 3 m M 5,98 ⋅ 1024 kg ⋅ 10 ⋅ 10246 kg m 7,57⋅ 10 7,3 5,98 3 ⋅ m6 −10 11 N T 2 r ⋅ ⋅ π 2 π 7,57 10 m 24 v = 7 3 ⋅ m/s , s 6 v = G = 6 , 67 ⋅ 10 ⋅ 3 (Masa y radio de laRTierra: 5,98 ∙ 10m kg; 6 370 km) ⋅ 10 6 T = = = 6 , 5 ⋅ 10 s T = 6,37 2 3 kg r 7,57 ⋅ 10 24m h = 1,2 ⋅6 106 m vv = 7, 3 ⋅ 10−11 m/s 3 ⋅m M N m 2 5,98 ⋅ 10 kg ⋅ 10 T = 6,37 v = G T = 6 ,67 7,3 ⋅ 10⋅ 103 s ⋅ MT =R5,98 ⋅ 1024 kg m 6 2 7, 3 ⋅ 10 7,57 ⋅ 103 6 m kgm 2 πrr ⋅10 m/s 2vπ=⋅ 7,57 MT = 5,98 ⋅ 1024 kg6 6 T = = = 6 ,5 ⋅ 10 s R T = 6,3724⋅ 10 m 2 πr situar ⋅103 m/s 2 π ⋅ 7,57 m MTsituar = 5,98en⋅ 10 kgun satélite pe3. ¿Qué cuesta más, órbita 4. T¿Podemos 7, 3 ⋅310m = v = v = satélites = 6 ,5 ⋅ 103 s a 6 geoestacionarios 106
1m ⋅ 10⋅10 2 πr ⋅ 7,57 2 π7,3 sobre superficie 3lasm 2 6 T diferentes =−11 vN2⋅πmr=alturas = 6 ,5terrestre, ⋅ 103 s o 3 ⋅ 7,3 10 ⋅ 2 π 7,57 24 ⋅10 m v 2 π ⋅ 6,67 ⋅por 10T el 5 , 98 10 kg ⋅ ⋅ m contrario, esta altura es fija e invariable? = = 6 ,5 ⋅ 103 s s 2 = 7,3 ⋅ 10 3 3 m MT = 5,98 ⋅ 1024 kg kg 1 v 6 Justifica tu respuesta. v = 2 πrs 27,3 π ⋅ 7,57 ⋅ 10N ⋅⋅s10 3 3 m2 m 86 400
MT =tu 5,98 ⋅ 1024 kg sado o uno ligero? Justifica respuesta.
D
Calcula la velocidad orbital y la altura sobre el ecuador a la que debe situarse un satélite geoestacionario.
T = 2 π ⋅ 6,67 = ⋅10 −11 =⋅610 ,524⋅ 10 s ⋅ 5, 98 kg s v 3 2m kg ⋅ 7,3 10 v = 86 400ss 3
v = 3, 1 ⋅10 m/s
— Calculamos la velocidad N ⋅orbital: m2
1 31 13
400 s h = r −R T 3, 1 ⋅10 3 m/s calcular la altura av la=superficie de la Tierra, h: 1 Aplicamos las ecuaciones de la velocidad orbital 7 6 h = 4 ,26 ⋅ 10 m − 6, 37 ⋅10 m 3 Tv 2 π GM M T T y del período para obtener un siste; vv == de r = G revolución h = r −R3 m 7 T M M T dos incógnitas: m s ⋅ 3 1 ⋅ 10 T3 m h = 3, 62 86 ⋅10400 r T con 1 m ma2 πde dosv ecuaciones = G 7 6 Tv s37 vM= G ss⋅⋅10 =400 4 ,26 − 63,m ⋅10 ⋅3311m⋅⋅10 10 r = T vT v= 86h86400 = 4m, 26 ⋅ 107 m r TrTv ; vr = 2 π GM T 3 v = 2G 3 s rπ= s M 7 7 86 400 ⋅ 3 1 ⋅ 10 s 2 π 2 π 7 T = 2 πvrr2=π2 πGr T T r =rT=v = = =44, ,26 26⋅ 10 ⋅ 10 mm h = 3, 62 ⋅10 s3mm = 86 400 ⋅ 3 1 ⋅ 10 s 7 2 2 π π 2 2 π π T =2vπTr = r r = = = 4 26 ⋅ 10 , m — Primero T vdebemos hallar el radio de s la órbita para v 2r π= 2π T = v = = 4, 26 ⋅ 107 m 2 πr calcular de la Tierra, h: 2 πla altura a la superficie 2π v T = v h = r - RT Al despejar r de la segunda ecuación y sustituirla =7rm − R T ∙ 106 m h = 4,26 h ∙ 10 6,37 = r- − RT en la primera, obtenemos: 1 hh= 1 7 r −R7T 6 , m , 37 h = 4 26 ⋅ 10 − 66m ⋅10 7 ∙ 10 6m 3 1 π Tv 2 GM h = 3,62 T π GM , m , h = 4 26 ⋅ 10 − 37 ⋅ 10 3 T v 2 h = r − R 7 r = T v r ; =v = ; 2vπ GM h= r −−T6R,T37 ⋅106m , m m h = 4 26 ⋅ 10 3 T = 1 7 T 1 7m h h= 3, 3762 66 r =T2 vπ ; v2 π= 2 π T T3 7 ⋅10 , m = 62 ⋅ 10 , m , m h = 4 26 ⋅ 10 − 6 37 ⋅ 10 GM 2 π GM T 3 , 37 ⋅10 m h = h4 ,26 Tv = 3⋅,10 62 m ⋅10−76m r = 2 π ;r v= = ; v T= T 7 7 2π T 2π T 3, 62 , 62 h =h3= ⋅10⋅10m m 6. Un objeto lanzado desde una nave espacial que5. Calcula la velocidad orbital y el período de reda en órbita circular alrededor de la Tierra con volución de un satélite que describe órbitas de 8 una velocidad de 2,52 ∙ 104 km/h. Calcula: a. el 500 km de radio alrededor de la Tierra. radio de la órbita; b. el período de revolución.
Prohibida su reproducción
1023 ⋅m/s 2 π ⋅ 6,67 ⋅10 −v11= 3N, 1⋅⋅m 5, 98 ⋅ 10 24 kg −11 kg 2 ⋅ 5, 98 ⋅ 10 24 kg 313 22 2 v = 2 π ⋅ 6,67 ⋅10−11−11NNkg Solución ⋅m m ⋅ MT 24 10 kg 1 v = G 3 ⋅m 6,67 98 ⋅ 10 24 kg ⋅1086 4002 2s ⋅ ⋅55, ,98 = 2s π⋅ 23⋅ π6,67 86v400 1 ⋅⋅10 r kg kg 2 86 400 s Tv s 3 RESOLUCIÓN. 7 N⋅ 10 ⋅mm −,11 r = = v =v = 2 π ⋅ 6,67 ⋅10 = 486 26400 2 πr MT ⋅ 5, 98 ⋅ 10 24 kg 86 400 s 2 2 π π 2s 3 m T = v = G — Datos: Un satélite geoestacionario 86 400 s ⋅ 3 kg 1 ⋅ 10 v r debe tener un v =Tv v = 3, 1 ⋅10 3 m/s s período de revolución igual2 πalr de rotación de la 86 400 s r = = v = 3, 1 ⋅10 3 m/s = 4, 26 ⋅ 107 m 2π 2 π 33 = Tierra alrededor de su Tpropio eje. T = 24 h = 86 v = 3 , 1 ⋅ 10 m/s v v =hallar 3, 1 ⋅10 m/s de la órbita para — Primero debemos el radio
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