Finding the final speed of a space probe using work and kinetic energy

In summary, the work done by the probes is about 30x greater than the initial kinetic energy, but the final speed is only about 13% greater than the initial speed.
  • #1
aqryus
6
1
Homework Statement
A deep space probe of mass 4.55 x10^4kg is travelling at an initial speed of 1.22 x10^4
m/s. The engines of the probe exert a force of magnitude 3.85 x10^5
N over 2.45 x10^8 m. Determine the probes final speed
Relevant Equations
W=Ek2-Ek1
W=fd
W=(1/2)mv2^2-(1/2)mv1^2
First I found work:

W=(3.85x10^5)(2.45x10^8)
W= 9.43x10^13

Then used that for difference of kinetic energy:

9.43x10^13 = (1/2) (4.55x10^4)v2^2 - (1/2)(4.55x10^4)(1.22x10^4)^2
9.43x10^13 = (22750)v2^2 - 3.386x10^12
9.43x10^13 + 3.386x10^12 = (22750)v2^2
9.77x10^13 = 22750v2^2
9.77x10^13/22750 = v^2
squareroot 4.3x10^6 = v
6.55x10^4 = v

But the answer is 1.38x10^4 m/s? I'm not sure where I went wrong. Any help is appreciated
 
Last edited:
Physics news on Phys.org
  • #2
Are the exponents for the distance and force correct?
 
  • #3
Frabjous said:
Is the exponent for the distance or force correct?
definitely an 8 on the worksheet
 
  • #4
If the work is a factor of 100 smaller, you get their answer. Check the units also. Also check the force.
 
  • #5
aqryus said:
squareroot 4.3x10^6 = v
6.55x10^4 = v
I don't know about the calculations that brought you to the two last equations but the very last equation does not follow from the next to last equation. The square root of 4.3 is a bit more than 2 and the square root of 10^6 is 10^3. So the answer that follows should have been a bit more more than 2000. You need to redo the calculations.
 
  • #6
kuruman said:
I don't know about the calculations that brought you to the two last equations but the very last equation does not follow from the next to last equation. The square root of 4.3 is a bit more than 2 and the square root of 10^6 is 10^3. So the answer that follows should have been a bit more more than 2000. You need to redo the calculations.
The 106 is a typo. Should be 109.
 
  • #7
Something is not adding up here.

The 9.4x1013 J of work is roughly 30 times greater than the initial kinetic energy of 0.34x1013 J. So, we'd expect the final speed to be a factor of roughly sqrt(30) times the initial speed, as the OP got.

However, the quoted "correct answer" is only about 13% greater than the initial speed (1.38/1.22=1.13). Even if the probe started from rest, doing that much work would give a much greater final speed.

Either there's a typo in the given quantities, or whoever is providing the "correct answer" made an error.
 
Last edited:
  • Like
Likes nasu
  • #8
Redbelly98 said:
Something is not adding up here.

The 9.4x1013 J of work is roughly 30 times greater than the initial kinetic energy of 0.34x1013 J. So, we'd expect the final speed to be a factor of roughly sqrt(30) times the initial speed, as the OP got.

However, the quoted "correct answer" is only about 13% greater than the initial speed (1.38/1.22=1.13). Even if the probe started from rest, doing that much work would give a much greater final speed.

Either there's a typo in the given quantities, or whoever is providing the "correct answer" made an error.
As noted by @Frabjous in post #4, taking either the thrust or the distance over which it is applied down two orders of magnitude gives the book answer.
 

1. How do you calculate the final speed of a space probe using work and kinetic energy?

The final speed of a space probe can be calculated using the formula: vf = √(2W/m), where vf is the final speed, W is the work done on the probe, and m is the mass of the probe.

2. What is the relationship between work and kinetic energy in determining the final speed of a space probe?

Work and kinetic energy are directly related in determining the final speed of a space probe. The work done on the probe is equal to the change in its kinetic energy, which ultimately determines its final speed.

3. Can the final speed of a space probe be calculated without knowing the work done on it?

No, the final speed of a space probe cannot be calculated without knowing the work done on it. The work done is an essential component in the equation for calculating the final speed.

4. How does the mass of the space probe affect its final speed?

The mass of the space probe has an inverse relationship with its final speed. This means that a heavier probe will have a lower final speed compared to a lighter probe, assuming the same amount of work is done on both.

5. Are there any other factors that can affect the final speed of a space probe?

Yes, there are other factors that can affect the final speed of a space probe, such as external forces acting on the probe, the direction of the work done, and the initial speed of the probe. These factors must also be taken into consideration when calculating the final speed using work and kinetic energy.

Similar threads

  • Introductory Physics Homework Help
Replies
1
Views
708
  • Introductory Physics Homework Help
Replies
6
Views
1K
Replies
7
Views
263
  • Introductory Physics Homework Help
2
Replies
56
Views
1K
  • Introductory Physics Homework Help
Replies
6
Views
995
  • Introductory Physics Homework Help
Replies
3
Views
857
  • Introductory Physics Homework Help
Replies
29
Views
887
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
806
  • Introductory Physics Homework Help
Replies
6
Views
3K
Back
Top