Variation of gravitational attraction between Sun and Earth

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Homework Help Overview

The discussion revolves around the variation of gravitational attraction between the Sun and Earth, specifically examining the forces at perihelion and aphelion distances. Participants are analyzing the gravitational force using the formula F = G × Msun × MEarth / distance², with given values for distances and masses.

Discussion Character

  • Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the gravitational force at two different distances but questions the discrepancy between their results and the textbook answer. They explore whether the issue lies in logical or arithmetic errors. Other participants suggest checking the squaring of the distance and converting units from kilometers to meters.

Discussion Status

Some participants have provided guidance on unit conversion and the importance of squaring the distance in the calculations. There is an acknowledgment of differing interpretations regarding the calculations, but no explicit consensus has been reached.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. The original poster expresses confusion about the correctness of their calculations compared to a textbook reference.

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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