Variation of gravitational attraction between Sun and Earth

AI Thread Summary
The discussion focuses on calculating the gravitational attraction force between the Sun and Earth at perihelion and aphelion using Newton's law of universal gravitation. The initial calculations resulted in a force of 3.68×10^28 N at perihelion and 3.44×10^28 N at aphelion, leading to a difference of 2.4×10^27 N, which did not match the textbook answer of 2.37×10^21 N. A key error identified was the failure to square the distance in the formula and the need to convert kilometers to meters before squaring. After correcting these mistakes, the user successfully arrived at the correct answer. The discussion highlights the importance of unit conversion and proper application of mathematical operations in physics calculations.
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 ?
 
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Matthew 289 said:
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.
What if you were given the distances in miles, instead of km?
 
First of all, you haven't squared the distance in the solution.
I don't have a clue of the calculations.
And secondly, convert kilometers into meters and then square.
that means 10^(8) changes to 10^(11) first since 1 km = 10^3 metres
Then, squaring becomes 10^(22)
rest of the method is alright,
i have solved it. and got the correct answer.
peace!
 
Saurabh said:
First of all, you haven't squared the distance in the solution.
I don't have a clue of the calculations.
And secondly, convert kilometers into meters and then square.
that means 10^(8) changes to 10^(11) first since 1 km = 10^3 metres
Then, squaring becomes 10^(22)
rest of the method is alright,
i have solved it. and got the correct answer.
peace!
Thank you very much ! I've done what you've suggested and the answer was right !
Have a nice day.
 

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