Finding the Kinetic Energy using the mass, radius and initial KE

[Ella1777]

Information Given:Zero, a hypothetical planet, has a mass of 4.2 x 1023 kg, a radius of 2.8 x 106 m, and no atmosphere. A 10 kg space probe is to be launched vertically from its surface.

Question: (a) If the probe is launched with an initial kinetic energy of 5.0 x 107 J, what will be its kinetic energy when it is 4.0 x 106 m from the center of Zero? I only put part A because i got part B correct
here is my attempt:
(G*m1*m2/d1)-(G*M1*M2/d2)= 3.0015000*10^7 J
What am I doing wrong?
What are some other calculations?
Thank you!

 

[haruspex]

(G*m1*m2/d1)-(G*M1*M2/d2)= 3.0015000*10^7 J
What am I doing wrong?

Fine so far, except that you quote too many significant figures. But you have bit more work to do to arrive at the answer to the question asked.

 

[Dr Dr news]

Just apply the conservation of the total energy which in this case is the sum of the potential energy and the kinetic energy. You are already part way there.

 

[Ella1777]

Just apply the conservation of the total energy which in this case is the sum of the potential energy and the kinetic energy. You are already part way there.

So I just found kinetic energy only not the initial I’m assuming.

Adding KE and PE will get my initial KE?

Or do you mean
KEi+KEf=PEi+PEf ??
If that’s the case then did I just solve for KEf and how do I find PEi and PEf?

So I add KE: 3.0*10^7 to PE?
How would I solve for PE?
I’m sorry I’m just quite confused how to use this equation I started on and apply it to conservation of energy.

Thank you I appreciate your help and efforts!

 

[Ella1777]

Fine so far, except that you quote too many significant figures. But you have bit more work to do to arrive at the answer to the question asked.

When working out the problem I have to quote all the significant figure or else points will be taken and it’s a risk to the wrong answer my professor says. When it comes to the final answer I do the sig figs and I typed in 3.0e7. Significant figures are usually the least of my problems it really isn’t one at all compared to understanding the process. Thank you!

 

[haruspex]

So I just found kinetic energy

You did? Looks to me like you just found a difference between two potential energies. How does that relate to kinetic energies here?

 

[Dr Dr news]

The conservation of mechanical energy applies to the total mechanical energy in the absence of losses say to friction, that is E(total, final) = E(total, initial). You are getting closer to a solution.

 

[MartinCarr]

I have not tried it with a different zero PE at the centre, but outcome should be the same..

 

[Dr Dr news]

As you know, the position at which you assume the potential energy to be zero is arbitrary, however some zero locations make understanding the physics a little more straightforward. The usual procedure in space flight analysis is to set the potential energy equal to zero at an infinite distance from the center of a large mass. Which when you look at the expression for potential energy makes perfect sense. This potential energy function leads to a gravitational potential energy well which must be overcome by a sizable amount of work to escape from a large planet. As you have calculated, energies of the order of 10^7 Joules.

 

[haruspex]

I have not tried it with a different zero PE at the centre, but outcome should be the same..

View attachment 224668

I am a bit confused.. are Ella1777 and MartinCarr the same person? I shall assume so.

Here's your problem:
What you keep doing is
KE final = change in PE
That is not what the law of conservation of mechanical energy says. What does it say?

 

[Ella1777]

I am a bit confused.. are Ella1777 and MartinCarr the same person? I shall assume so.

Here's your problem:
What you keep doing is
KE final = change in PE
That is not what the law of conservation of mechanical energy says. What does it say?

I’m not Martin Carr, as for I don’t understand anything of what he did if anything he confused me because I know conservation of energy must take place and I know what it looks like and he simply did not do that.
I’m Ella1777 and I asked the question and he attempted to give me a solution that is incorrect and confuse me more unfortunately.
My professor simply explained to me(I just asked 5 minutes ago) that I do
Ugrav (o) +K(o)=Ugrav(t)+K(t)
We plug in everything else and solve for K(t)
So following what my professor said I attempted
-(6.67e-11)*((4.2e23)(10)/(2.8e6)) +5.0e7= -(6.67e-11)*((4.2e23)(10/(4.0e6))+K(t)
Then I solve for K(t)
But when I solved for K(t) I got 0.7146426787
Which I think it’s incorrect. Am I plugging in a wrong number or is it my algebra?
Or is it that I solved t and not K (then how would find K?)

 

[Dr Dr news]

Your equation is correct and the numbers are correct. You must not have punched them into the calculator correctly. You had the difference in the potential energies calculated correctly in post #1. It's just realizing that you are going from a high negative potential energy at the bottom of the gravitational well and a high positive kinetic energy (it is always positive) to a region with a lower negative potential energy as you move up the gravitational well and a lower positive kinetic energy. The sum, the total energy, is a constant. The difference in kinetic energy from start to finish represents the work done in traversing against the mutual attraction of the two masses.

 

[haruspex]

Ugrav (o) +K(o)=Ugrav(t)+K(t)
We plug in everything else and solve for K(t)
So following what my professor said I attempted
-(6.67e-11)*((4.2e23)(10)/(2.8e6)) +5.0e7= -(6.67e-11)*((4.2e23)(10/(4.0e6))+K(t)
Then I solve for K(t)
But when I solved for K(t) I got 0.7146426787

As Dr Dr news writes, that is all correct except for the final answer.
But you were very close in your post #1. There you correctly found the gain in PE. You just had one simple step to do to reach the answer.
Can you write what you did in post #1 in terms of the symbols in the equation Ugrav (o) +K(o)=Ugrav(t)+K(t)?