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Matthew 289
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Homework Statement
Distance Earth-Sun at perihelion = 1.471×108 km
Distance Earth-Sun at aphelion = 1.521×108 km
Sun mass = 2.0×1030 kg
Earth mass = 5.972×1024 kg
G = 6.67×10-11 m3/kg⋅s
What is the change in Newton of the attraction force between the Sun and the Earth from the perihelion to the aphelion ?
Homework Equations
F= G × Msun×MEarth/distance2
The Attempt at a Solution
Fperihelion= (6.67×10-11 m3/kg⋅s) × (2.0×1030 kg)×(5.972×1024 kg)/(1.471×108 km) = 3.68×1028 N
Faphelion= (6.67×10-11 m3/kg⋅s) × (2.0×1030 kg)×(5.972×1024 kg)/(1.521×108 km) = 3.44×1028 N
ΔF= |Fperihelion| - |Faphelion| = 3.68×1028 N - 3.44×1028 N = 2.4×1027 N
My problem:
The result does not correspond to that given by the textbook (2.37×1021 N).
I cannot see the fault in my resolution and other methods I've tried give wrong answers as well.
Can anyone see my mistake/s ? Is it logical or arithmetical ?