Variation of gravitational attraction between Sun and Earth

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SUMMARY

The variation of gravitational attraction between the Sun and Earth is calculated using Newton's law of universal gravitation. The force at perihelion is determined to be 3.68×1028 N, while at aphelion it is 3.44×1028 N, resulting in a change of 2.4×1027 N. A common error identified in the discussion is the failure to square the distance when applying the formula F = G × MSun × MEarth / distance2. Correcting the distance from kilometers to meters and squaring it resolves discrepancies with textbook values.

PREREQUISITES
  • Understanding of Newton's law of universal gravitation
  • Familiarity with unit conversion (kilometers to meters)
  • Basic algebra for squaring numbers
  • Knowledge of gravitational constant (G = 6.67×10-11 m3/kg⋅s2)
NEXT STEPS
  • Review calculations involving gravitational force using F = G × M1 × M2 / r2
  • Practice unit conversions, specifically between kilometers and meters
  • Explore the implications of gravitational variations in celestial mechanics
  • Learn about the significance of perihelion and aphelion in orbital dynamics
USEFUL FOR

Students studying physics, particularly those focusing on gravitational forces and celestial mechanics, as well as educators seeking to clarify common calculation errors in gravitational attraction problems.

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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