Gravitational Formulas for Solving Energy-Based Questions

  • Thread starter Thread starter Clara Chung
  • Start date Start date
  • Tags Tags
    Gravitational
Click For Summary

Homework Help Overview

The discussion revolves around gravitational formulas and energy conservation principles related to escape velocity and the motion of a spacecraft. Participants are analyzing the relationships between kinetic energy, potential energy, and gravitational forces in the context of a specific problem involving a planet's mass and escape conditions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants attempt to derive escape velocity using energy conservation principles but express confusion regarding the application of formulas meant for circular orbits. Questions arise about the definitions of variables and the correct approach to determine the planet's mass.

Discussion Status

Some participants have suggested alternative methods, including graphical approaches and energy conservation equations, while others are questioning the assumptions made in the original attempts. There is an ongoing exploration of different interpretations and methods without a clear consensus on the correct approach.

Contextual Notes

Participants note the importance of defining variables clearly and the distinction between scenarios involving circular orbits versus escape trajectories. There is also mention of potential discrepancies in provided answers and the need for careful consideration of energy conservation principles.

Clara Chung
Messages
300
Reaction score
13

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
 

Attachments

  • grav.png
    grav.png
    56.1 KB · Views: 488
Physics news on Phys.org
Clara Chung said:

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 .

Clara Chung said:
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.
 
  • Like
Likes   Reactions: Clara Chung
ehild said:
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?
 
Clara Chung said:
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.
 
Last edited:
  • Like
Likes   Reactions: Clara Chung
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.
 
  • Like
Likes   Reactions: Clara Chung
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.
 
  • Like
Likes   Reactions: Biker and 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
 
Clara Chung said:
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.
 
  • Like
Likes   Reactions: Clara Chung

Similar threads

Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
1K
Replies
6
Views
822
  • · Replies 25 ·
Replies
25
Views
4K
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
4
Views
1K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 39 ·
2
Replies
39
Views
4K