Calculating Changes in Earth-Sun Distance with Varying Sun Mass

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The discussion focuses on calculating the change in Earth's distance from the Sun if the Sun's mass is reduced by 15%, while maintaining angular momentum. The gravitational and centrifugal forces are equated to derive the new distance, resulting in Earth being closer to the Sun at 0.85 AU compared to the current 1 AU. A participant expresses confusion over why a lighter Sun would result in a closer Earth, suggesting that reduced gravitational effects should increase distance. Clarification is provided that the conservation of angular momentum dictates that Earth's speed and distance must adjust accordingly. The conversation concludes with an acknowledgment of understanding the calculations involved.
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Homework Statement


Calculate the relative change in the distance of the Earth from the Sun, if the mass of the Sun is 15% lower than today weight of the Sun. Suppose that the Earth moves and will move along a circular path and will maintain its angular momentum.


Homework Equations


Equality of gravitational and centrifugal forces.


The Attempt at a Solution


For mass of the Sun today:
Valid for Earth: The centrifugal force = gravitational force
m_e*(v)^2/r=κ*m_e*m_s/(r)^2
(v)^2=κ*m_s/r


For mass of the Sun when is 15% lower than today weight of the Sun:
Valid for Earth: The centrifugal force = gravitational force
Mass and velocity of the Earth does not change because it does not change angular momentum.
m_e*(v)^2/r_1=κ*0,85m_s*m_e*/(r_1)^2
(v)^2=κ*0,85m_s/r_1

Equal squares of velocities:
κ*m_s/r=κ*0,85m_s/r_1
1/r=0,85/r_1

When is r=1 AU, then r_1=0,85AU

Why? I think it's stupid. Why the Earth is closer to the Sun, when the Sun is lighter? I think must be the Earth from the Sun farther, because the gravitational effects diminish. Why my ideas don't coincide with the results? How would you counted it?

Thanks very much.
 
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Numeriprimi said:

Homework Statement


Calculate the relative change in the distance of the Earth from the Sun, if the mass of the Sun is 15% lower than today weight of the Sun. Suppose that the Earth moves and will move along a circular path and will maintain its angular momentum.


Homework Equations


Equality of gravitational and centrifugal forces.


The Attempt at a Solution


For mass of the Sun today:
Valid for Earth: The centrifugal force = gravitational force
m_e*(v)^2/r=κ*m_e*m_s/(r)^2
(v)^2=κ*m_s/r


For mass of the Sun when is 15% lower than today weight of the Sun:
Valid for Earth: The centrifugal force = gravitational force
Mass and velocity of the Earth does not change because it does not change angular momentum.
m_e*(v)^2/r_1=κ*0,85m_s*m_e*/(r_1)^2
(v)^2=κ*0,85m_s/r_1

Equal squares of velocities:
κ*m_s/r=κ*0,85m_s/r_1
1/r=0,85/r_1

When is r=1 AU, then r_1=0,85AU

Why? I think it's stupid. Why the Earth is closer to the Sun, when the Sun is lighter? I think must be the Earth from the Sun farther, because the gravitational effects diminish. Why my ideas don't coincide with the results? How would you counted it?

Thanks very much.

The angular momentum is mvr. The product of the radius and speed is unchanged.


ehild
 
Sorry, but I do not understand you too much. Please, could you explain to me what to fix to make it right in great detail?
 
The speed of Earth (v) changes. Conservation of angular momentum means that mer v is conserved.

ehild
 
Well, I get it, thank you very much :-)
 

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