Answer on Question #61340-Physics-Mechanics-Relativity
A satellite of mass 2500 kg is orbiting the Earth in an elliptical orbit. At the farthest point from the Earth, its
altitude is 3600 km, while at the nearest point, it is 1100 km. Calculate the energy and angular momentum
of the satellite and its speed at the aphelion and perihelion.
Solution
We know, 𝑈 = −
𝐺𝑀𝑚
𝑟
(where G IS gravitational constant, m is satellite's mass, M IS Earth’s mass)
At aphelion,
𝑟1 = 𝑅𝐸𝐴𝑅𝑇𝐻 + 3600 𝑘𝑚
At perihelion,
𝑟2 = 𝑅𝐸𝐴𝑅𝑇𝐻 + 1100 𝑘𝑚
From the conservation of energy:
𝐾 + 𝑈 = 𝑐𝑜𝑛𝑠𝑡 →
𝑚𝑣12 𝐺𝑀𝑚 𝑚𝑣22 𝐺𝑀𝑚
−
=
−
2
𝑟1
2
𝑟2
From the conservation of angular momentum:
𝑚𝑣1 𝑟1 = 𝑚𝑣2 𝑟2
𝑣2 =
𝑟1
𝑣
𝑟2 1
2
𝑣12 𝐺𝑀 1 𝑟1
𝐺𝑀
−
= ( 𝑣1 ) −
2
𝑟1
2 𝑟2
𝑟2
1 1
(𝑟 − 𝑟 )
2
𝑣12 = 𝐺𝑀 1
𝑟1 2
1 − (𝑟 )
2
1. Speed.
At aphelion,
𝑣1 = √6.673 ∙ 10−11 ∙ 5.98 ∙ 10
1
1
(
6 + 3.6 ∙ 106 ) − (6.37 ∙ 106 + 1.1 ∙ 106 ))
(6.37
∙
10
24
106
2
106 )
(6.37 ∙
+ 3.6 ∙
1−(
)
(6.37 ∙ 106 + 1.1 ∙ 106 )
At perihelion,
𝑣2 =
2. The energy.
At aphelion,
(6.37 ∙ 106 + 3.6 ∙ 106 )
𝑚
4140.5 = 5526.2
6
6
(6.37 ∙ 10 + 1.1 ∙ 10 )
𝑠
= 4140.5
𝑚
𝑠
1
6.673 ∙ 10−11 ∙ 5.98 ∙ 1024 ∙ 2500
𝐸𝑎 = 2500(4140.5)2 −
= −7.86 ∙ 1010 𝐽.
(6.37 ∙ 106 + 3.6 ∙ 106 )
2
At perihelion,
𝐸𝑝 = 𝐸𝑎 = −7.86 ∙ 1010 𝐽.
2. The angular momentum.
At aphelion,
𝐿𝑎 = 2500(4140.5)(6.37 ∙ 106 + 3.6 ∙ 106 ) = 1.03 ∙ 1014
At perihelion,
𝐿𝑝 = 𝐿𝑎 = 1.03 ∙ 1014
𝑘𝑔𝑚2
.
𝑠
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𝑘𝑔𝑚2
.
𝑠