Comparing gravitational force at varying distances

[KLI897]

Homework Statement

Halley’s Comet orbits the sun about every 75 years due to the gravitational force the sun provides. Compare the gravitational force between Halley’s Comet and the sun when the comet is at aphelion (its greatest distance from the sun) and d is about 4.5 x 10^12 m to the force at perihelion (or closest approach), where d is about 5.0 x 10^10 m.

Homework Equations

F=GmM/d^2
(Not applicable because masses aren't provided?)

The Attempt at a Solution

I wasn't sure where to begin considering the masses were not provided, but eventually I found that the distance from the sun at the furthest distance is 90x greater than when it is at its closest point. I then assumed that it wanted to me to use Newton's Law of Universal Gravitation that as the distance is doubled the force decreases by 1/4. But I'm not sure if where to go from here.

 

[ehild]

Homework Statement

Halley’s Comet orbits the sun about every 75 years due to the gravitational force the sun provides. Compare the gravitational force between Halley’s Comet and the sun when the comet is at aphelion (its greatest distance from the sun) and d is about 4.5 x 10^12 m to the force at perihelion (or closest approach), where d is about 5.0 x 10^10 m.

Homework Equations

F=GmM/d^2
(Not applicable because masses aren't provided?)

The Attempt at a Solution

I wasn't sure where to begin considering the masses were not provided, but eventually I found that the distance from the sun at the furthest distance is 90x greater than when it is at its closest point. I then assumed that it wanted to me to use Newton's Law of Universal Gravitation that as the distance is doubled the force decreases by 1/4. But I'm not sure if where to go from here.

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The problem asks the ratio of the forces, F(aphelion)/F(perihelion). Yes, use the Law of Universal Gravitation.

 

[KLI897]

.

The problem asks the ratio of the forces, F(aphelion)/F(perihelion). Yes, use the Law of Universal Gravitation.

Thank you! I've concluded that the force of the sun's pull at aphelion is only 1/8100 of whatever it is at perihelion by using the inverse-square law . It was much simpler than I originally thought, I must've been over thinking it.