Potential energy turned into kinetic

[cragar]

Homework Statement

We have an object with much smaller mass than compared to the sun.
This object is at rest when released at 1AU from the sun. After a week how far is the mass from the sun.

The Attempt at a Solution

So I look at the gravitational potential energy and set it equal to kinetic energy.
$\frac{-GMm}{r}=\frac{m(v)^2}{2}$
the v should be an r dot. now I take the square root of both sides. and then multiply both sides by dt then move the r from the left side to the right side and then integrate bothe sides. this will give me r(t). and then for the bounds I use r=1au and then I can solve for r final. after using t=0 and t=1 week for the time. i will put the time in seconds. this seems somewhat reasonable.

 

[Orodruin]

Why would the kinetic energy equal the gravitational potential? There is nothing saying this has to be the case (furthermore, the potential is only defined up to an arbitrary constant) and in particular it does not make sense in your case since the potential is strictly negative and kinetic energy non negative.

You can go one of two ways:
1. Write down the force equation and solve for r as a function of time.
2. Use conservation of energy (this will save you one integration)

 

[cragar]

for the force equation would I use F=ma and set it equal to the gravitational force.
For conservation of energy would I say that kinetic plus potential is a constant.

 

[Orodruin]

Yes, both would be correct approaches so I suggest you try doing either of those and return with the result. For the second approach, remember that $v = -\dot r$ and (for both approaches) that $v(t=0) = 0$ and $r(t=0) = R_\oplus$.

 

[HalfLostAndSearching]

Your approach to solving for the distance of the object from the sun after one week is correct. By equating the gravitational potential energy to the kinetic energy and then solving for the distance using integration, you can determine the final distance of the object from the sun after one week. It is important to note that this approach assumes that there are no other forces acting on the object, such as air resistance or the gravitational pull of other bodies.

Additionally, it is worth mentioning that the object's mass and initial distance from the sun will also affect its final distance after one week. A smaller mass and larger initial distance will result in a larger final distance, while a larger mass and smaller initial distance will result in a smaller final distance. It may be helpful to consider these factors when solving for the final distance.

Overall, your approach is scientifically sound and reasonable. Keep in mind that in real-world scenarios, there may be other factors at play that could affect the final distance of the object from the sun.