Calculating Changes in Earth-Sun Distance with Varying Sun Mass

[Numeriprimi]

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.

 

[ehild]

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

 

[Numeriprimi]

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?

 

[ehild]

The speed of Earth (v) changes. Conservation of angular momentum means that m_er v is conserved.

ehild

 

[Numeriprimi]

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