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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

El radio de la órbita es igual al radio de la Tierra más
COMPRENSIÓN. r =R T +h =6,37 ⋅ 10 m +1,2 ⋅ 10 m =7,57 ⋅ 10 m
6
6
6
la distancia del satélite a la superficie terrestre:
m =5000 kg
6
6
6
r =R T +h =6,37 6 ⋅ 10 m +1,2 ⋅ 10 m =7,57 ⋅ 10 m
6
6
r = RT + h = 6,37 ∙ 10 m + 1,2 ∙ 10 m = 7,57 ∙ 10 m
6
6
6
24
5,98 +1,2
6
h =1,2 ⋅ 10 m m =5000 kg r =R +h =6,37 ⋅ 2 10 m ⋅10 ⋅ 10 m =7,57 ⋅ 10 m
a. Calculamos la velocidad orbital del satélite:
kg
T
M
−11 N
⋅m
6
=
,
6 67
⋅6
6
⋅ 10
v = G T r =R +h =6,37 2 ⋅ 10 m +1,2 ⋅ 10 m =7,57 ⋅ 10 m
6 6
m =5000 kg r T kg 7,57 ⋅10 m
6
6
24
6
6
R T =6,37 ⋅ 10 m h =1,2 ⋅ 10 m r =R +h =6,37 ⋅ 10 m +1,2 ⋅ 10 6 2m =7,57 ⋅ 10 m
⋅10
m =5000 kg
kg
T
5,98
M
−11 N
⋅m
v
v = 7 3 , G= ⋅10 3 T m/s 6 67 10, ⋅ ⋅
=
m =5000 kg r 2 7,57 6 6
6 m
⋅1024
6
6
kg 2
6
h =1,2 ⋅ 10 m r =R T +h =6,37 ⋅ 10 m +1,2 ⋅ 10 m =7,57 ⋅ 10 m
5,98
⋅10
kg
⋅m
M
N
6
R T =6,37 ⋅ 10 m T −11
,
6
h =1,2 ⋅ 10 m v = G = 6 67 ⋅ 10 3 2 ⋅ 24 kg
M T =5,98 ⋅ 10 24 kg m =5000 kg M T 6 6 67 10, v = 7 3 , −11 ⋅10 ⋅N m 2 m/s 5,98 ⋅10 6 6 6
r
⋅10
6 7,57
24 m
⋅
6
2 ⋅
G
=
kg
6
⋅10
5,98
kg
2π
h =1,2 ⋅ 10 m v = r =R πT +h =6,37 ⋅ 10 m +1,2 ⋅ 10 m =7,57 ⋅ 10 m
⋅10 m
r
⋅ 7,57
2
6 6
M
N
3 ⋅m2
T r
6
= 6 5
=
R T =6,37 ⋅ 10 m T = b. Calculamos el período de revolución: 7,57 ⋅10 m
, −11
⋅10 s kg
,
⋅ 10
6 67
m =5000 kg v = G = m v = 7 3 , 3 ⋅
v
2
6
R T =6,37 ⋅ 10 m 6 24 kg r 7,3 M ⋅10 3 ⋅10 m/s 2 5,98 ⋅1 24 6 6 kg
⋅100 m
7,57
M T =5,98
h =1,2 ⋅ 10 m ⋅ 10
−11 N kg
⋅m6
T
π
(Masa y radio de la Tierra: 5,98 ∙ 10 kg; 6 370 km) v = G T = = r2π v = = ⋅ ⋅ 7,57 m/s⋅3 10 3 ⋅10 ⋅ m = 6 5 ⋅10 3 6 6 s
s 6 67 10, 2 7,
6
24
R T =6,37 ⋅ 10 m
,
2
6
−11 N 3 m/s
h =1,2 ⋅ 10 m r M v v = 7 3 , ⋅10 3 kg ⋅m 2 7,57 ⋅10 24 m kg
⋅10
5,98
m
6
R T =6,37 ⋅ 10 m T 7,3 ⋅10
6 67
,
M T =5,98 ⋅ 10 24 kg v = G = v = 7 3 , ⋅ 10 3 m/s ⋅ 6 6
s
6
2
⋅10
M T =5,98 ⋅ 10 24 kg 2π r r 2 π ⋅ 7,57 ⋅10 kg m 7,57 ⋅10 m
3
,
6
R T =6,37 ⋅ 10 m T = = 6 = 6 5 ⋅10 s
3 m
⋅ 7,57
π
⋅10
¿Podemos situar satélites geoestacionarios a
3. ¿Qué cuesta más, situar en órbita un satélite pe- 4. T = 2π r = 2 v = 7 3 , ⋅10 m m/s = 6 5 ⋅10 s
24
3
M T =5,98 ⋅ 10
kg
v
,
3
6
π 7,3
⋅10
2π
2
sado o uno ligero? Justifica tu respuesta. diferentes alturas sobre la superficie terrestre, o
r
⋅10
⋅ 7,57
1m
m
24
v
3
kg
M T =5,98 ⋅ 10
T = N ⋅m r 2 = ⋅10 3 s 3 6 = 6 5 ⋅10 s
,
π,3
24 ⋅10 m
−11 2π
por el contrario, esta altura es fija e invariable?
kg m
3
⋅10
2π ⋅ 6,67 ⋅10 T = v ⋅ 5 98, 27 ⋅ 7,57 3 s = 6 5 ⋅10 s
,
2 =
⋅10
7,3
M T =5,98 ⋅ 10 24 kg Justifica tu respuesta. m 1
kg v
v = 2π r 27 π ⋅,3 10 3 ⋅10 m
6
⋅ 7,57 s
2
3 3
86 400
,
T = s = −11 N ⋅m = 6 5 24 ⋅10 s
2π ⋅ 6,67 ⋅10 s 2 m ⋅ 5 98, ⋅10 kg
v
D v = 7,3 ⋅10 3 kg
3
v = 3 1 ⋅10 m/s 86 400 s s
,
1
Calcula la velocidad orbital y la altura sobre el ecuador — Calculamos la velocidad orbital: 3 1
2
N
⋅m
3
a la que debe situarse un satélite geoestacionario. 2π ⋅ 6,67 ⋅10 −11 = 3 1 ⋅m 2⋅ 5 ⋅10 24 kg 1 3
⋅10 m/s 98,
v
,
N 2
2π ⋅ 6,67 ⋅10 −11 kg 2 ⋅ 5 98, ⋅10 24 kg 1
Solución v = 2 3 3
N ⋅m 2
N
⋅m
M −11 −11 kg 24 24
2π ⋅ 6,67
⋅ 6,67 ⋅10
2π
v = G T 86 400 s ⋅ 3 1 ⋅10 3 m ⋅1086 400 2 ⋅ s ⋅ 5 98, 5 98, ⋅10 kg kg 1
⋅10
v =
2
kg
kg
RESOLUCIÓN. r T v v = s 86 400 7 m 2 s 3
⋅
N
, −11
2π r M r = = v = 2π ⋅ 6,67 = 4 26 ⋅10 m ⋅ 5 98, ⋅10 24 kg
s
⋅10 86 400s86 400
T
T =
— Datos: Un satélite geoestacionario debe tener un 2 π 86 400 s ⋅ 3 3 2 3 m
2π
G
v =
kg ⋅1 10
v
r
,
período de revolución igual al de rotación de la v = T v v = 3 1 ⋅10 m/s s 7
3
,
2π r r = = v = 3 1 86 400 s = 4 26 ⋅10 m
⋅10 m/s
,
T =
Tierra alrededor de su propio eje. T = 24 h = 86 2π v = 3 1 2 π 3 3
,
⋅10 m/s
v v = 3 1 ⋅10 m/s
,
400 s — Primero debemos hallar el radio de la órbita para
h =
r −
R
3
,
⋅10 m/s
= 3 1
v
T
Aplicamos las ecuaciones de la velocidad orbital calcular la altura a la superficie de la Tierra, h:
1
7
6
GM M
,
⋅
,
T v
y del período de revolución para obtener un siste- h = 4 26 10 m −6 37 ⋅10 m
2π
3
T T
v = =
r = ; v G h = r − 3 m
R
T M
ma de dos ecuaciones con dos incógnitas: T v , 86 7 400 s ⋅ 3 1 7 ⋅10 T 3 3 m s m 6
2π
r
M
⋅10 m
T
h = 3 62
v =
G
1
T
G
v =
,
h =
⋅106 37
7
= 4 26
=
r
G r
v = 2π M T v r T ; v = 2π GM T 3 r = T v T v 86 86 400 s⋅4 26 1 ⋅400 s ⋅ 3 1 −0 m ⋅3 1 10 3 , m s s ⋅10 m , ⋅10 m
r =
T = 2π M T T 2π 86 400 s 2 π ⋅10 7 7 7
⋅ 3 1
3 m = 4 26 ⋅4 26 10 m
2π v = 2π G r r = r = = = h = 3 62 ⋅10 m = , , ⋅10 m
r r
,
2
T = v T = r T v 2π 86 400 s 2 π π ⋅ 3 1 ⋅10 s 7
2π
,
⋅10 m
=
= 4 26
r =
2π r v — Primero debemos hallar el radio de la órbita para
T v
s
v
7
r = π
2 π
2
,
T = 2π r calcular la altura a la superficie de la Tierra, h: ⋅10 m
=
= 4 26
v T = 2π 2 π
v
h = r - R
Al despejar r de la segunda ecuación y sustituirla h = R T
7 r −
6
T
en la primera, obtenemos: 1 h = 4,26 ∙ 10 m - 6,37 ∙ 10 m
r −
h =
R
T
7
1 h = r − R 7 T 6
T v 2π GM 3 1 h = 4 26 10 m ⋅10 6m
h = 3,62 ∙ 10 m 37
,
7 −6,
⋅
⋅ h
,
r −
−6 37
T v
T
;
r = T v r = v = ; 2π 2π GM T 3 h = 4 26 10 m r − T R , T 37 ⋅10 m Prohibida su reproducción
6
h = 4 26 10 m R=7
h =
v = GM
⋅
⋅10 m
−6,
,
1 3
7
7 m
⋅10
r = 2π ; v 2π= T T T 1 h = 3, 7 62 7 ⋅2 10 m 6 6
h = 3 6 ,
, ⋅
,
−6 37 ⋅37 10 m
T v T v 2π GM T 2π 3 GM 3 h = h = 4 26 10 m −4 26 10 m 6,7 , ⋅10 m
⋅
2π
T
,
r = ; r = v = ; v = T h = 3 62 ⋅10 m
7
7
2π 2π T T h = , , ⋅10 m
h = 3 62 ⋅3 62 10 m
5. Calcula la velocidad orbital y el período de re- 6. Un objeto lanzado desde una nave espacial que-
volución de un satélite que describe órbitas de 8 da en órbita circular alrededor de la Tierra con
500 km de radio alrededor de la Tierra. una velocidad de 2,52 ∙ 10 km/h. Calcula: a. el
4
radio de la órbita; b. el período de revolución.
77