Gravitational Formulas for Solving Energy-Based Questions

[Clara Chung]

Homework Statement

The problem is attached

Homework Equations

Energy, gravitational formulas

The Attempt at a Solution

For part b, the answer is 2km s^-1
If it can escape, KE-PE(energy traveled from infinity to R)=0
KE = PE
mv^2 /2 = GMm/R
v^2/2=GM/R..(1)

I tried to find GM by using the information provided
mv^2 /R = GMm /R^2
v^2/2 = GM/2R...(2)

which combining (1) and (2) I get v^2/2 = v^2 which is impossible.
Please tell me what's wrong and the correct method to solve the question

 

[ehild]

Homework Statement

The problem is attached

Homework Equations

Energy, gravitational formulas

The Attempt at a Solution

For part b, the answer is 2km s^-1
If it can escape, KE-PE(energy traveled from infinity to R)=0
KE = PE

It is not true if the spacecraft has kinetic energy at infinity .

mv^2 /2 = GMm/R
v^2/2=GM/R..(1)

I tried to find GM by using the information provided
mv^2 /R = GMm /R^2
v^2/2 = GM/2R...(2)

which combining (1) and (2) I get v^2/2 = v^2 which is impossible.
Please tell me what's wrong and the correct method to solve the question

How do you intend to determine the mass of the planet? You have to define what R and v are. It seems you try to use the formula valid for a circular orbit, but this spacecraft is leaving the planet instead of orbiting around it.

 

[Clara Chung]

It is not true if the spacecraft has kinetic energy at infinity .How do you intend to determine the mass of the planet? You have to define what R and v are. It seems you try to use the formula valid for a circular orbit, but this spacecraft is leaving the planet instead of orbiting around it.

then what formula should I use?

 

[cnh1995]

then what formula should I use?

I'm not sure this will work but here's what I think..
You can first plot the graph of v vs r on a graph paper. Draw a smooth curve. At (r=1, v=6.635), draw a tangent to that curve whose slope will be dv/dr.
Now,
deceleration=dv/dt=(dv/dr)*(dr/dt)=v*dv/dr..
You can know v, r and dv/dr at every point on the graph.
You can equate the deceleration obtained from graph to GM/r² and get M.

 

[Biker]

Hello,

I have searched a little bit about the topic of escape velocity and I figured the solution
I will give you an example
If you have a road that is 8 meters tall and you know that a section from the end to a point in that road is 5 meters tall. How far is the point from the starting point of the road? You just have to do subtraction.

Okay so If G M m /r gives you potential from a point to infinity and you take another (r value) above the first one you took and substract the two what do you get?

Now use energy conservation using velocities and find out GM.

Once you find that out we can continue.

 

[gneill]

Look at energy conservation. The total specific mechanical energy (that is, the energy per unit mass of the object in orbit) is the sum of the specific kinetic and potential energy at any instant. For a body in a free-fall trajectory of any kind (such as this spacecraft is) this value is a constant.

In particular: $\xi = \frac{v^2}{2} - \frac{GM}{r}$ is a constant over the whole trajectory.

You should be able to use this along with the given data to solve for both $\xi$ and GM.

 

[Clara Chung]

Thanks, I can find the answer of part a by using both methods.
The answer of part c is 6 x 10^23 kg, but when I use U^2/2 - GM/R =V^2/2, I found out 3 x 10^23 kg

 

[gneill]

Thanks, I can find the answer of part a by using both methods.
The answer of part c is 6 x 10^23 kg, but when I use U^2/2 - GM/R =V^2/2, I found out 3 x 10^23 kg

Your answer (3.0 x 10^23 kg) looks good to me. Perhaps there's an error in the given answer key.