Time required to change the velocity of an asteroid?

[Howard Fox]

Homework Statement

So I am given a problem where I am asked to find the time it takes to change the velocity of an asteroid by 1m/s. In a previous section of the problem I was asked to find the force through which the asteroid is being propelled, which I calculated it to be F=1.3x10^4 N, and the magnitude of the acceleration, being a=6.2x10^-5 m/s^2. I am also given the mass of the asteroid: 7.5x10^12kg.

2. Relevant Equations
Vf=Vo+a*t.

The Attempt at a Solution

I was thinking of using one of the basic kinematic equations, like the above, and to solve for time by imputting two different values of Vo and Vf. The problem is I don't know the initial speed and I am not sure I could find it with the given information.

Thank you for your help.

 

[phyzguy]

To find the change in velocity, do you need to know V0 and Vf? What does the "change in velocity" mean?

 

[jbriggs444]

F=1.3x10^4, and the magnitude of the acceleration, being a=6.2x10^-5. I am also given the mass of the asteroid: 7.5x10^12.

F=ma. At least one of those numbers must be incorrect. Technically, all of them are incorrect since none of them have units specified.

 

[Howard Fox]

To find the change in velocity, do you need to know V0 and Vf? What does the "change in velocity" mean?

It means change in position with respect to time?

 

[gneill]

Welcome to Physics Forums!

Please don't delete the template headers.

You should specify the units associated with your values. A number alone is practically meaningless.

 

[Howard Fox]

F=ma. At least one of those numbers must be incorrect. Technically, all of them are incorrect since none of them have units specified.

Okay I have corrected the missing units. I could have previously made wrong calculations which resulted in the wrong numbers. I was asked to find the magnitude of the resultant force of the vectors of the two asteroid's thrusters, which supply a force with x and y components of:

I calculated this magnitude as 1.3x10^4 N. Is that correct?
Then the magnitude of the acceleration produced by the thrusters I calculated as:

Is there something I did wrong? Thank you very much for your help

 

[jbriggs444]

[I've re-rendered your attachments using LaTeX]

Okay I have corrected the missing units. I could have previously made wrong calculations which resulted in the wrong numbers. I was asked to find the magnitude of the resultant force of the vectors of the two asteroid's thrusters, which supply a force with x and y components of:
$F_1=(8.0\times 10^7, 7.0 \times 10^7)$
$F_2=(4.0\times 10^7, -2.0 \times 10^7)$
I calculated this magnitude as 1.3x10^4 N. Is that correct?

No. Can you show your work for this part? [I think I see what you did -- but we need you to show it explicitly. Hint: can you show us the Pythagorean formula?]

Then the magnitude of the acceleration produced by the thrusters I calculated as:
$a=\frac{F}{m}=\frac{1}{7.5 \times 10^{12}} \times (12 \times 10^7, 5 \times 10^7)$
$a=(1.6 \times 10^{-5}, 6.0 \times 10^{-5} )$
Is there something I did wrong?

This one contains a different error.

 

[Howard Fox]

[I've re-rendered your attachments using LaTeX]

No. Can you show your work for this part? [I think I see what you did -- but we need you to show it explicitly. Hint: can you show us the Pythagorean formula?]

This one contains a different error.

Okay, here is the work, sorry if the final answer is missing units again, it's in Newtons of couse.

 

[jbriggs444]

[Transcribing to LaTeX again]

Okay, here is the work, sorry if the final answer is missing units again, it's in Newtons of couse.
$$|F| = \sqrt{F_x^2 + f_y^2} = \sqrt{12 \times 10^7 + 5 \times 10^7} = 1.3 \times 10^4$$

The first error is on that line. What happened to the squaring that was supposed to be done on the x and y components?

Edit: Sanity checks are a useful tool. As I alluded to previously, if a force has an x component of 1.2 x 10⁸ then it should have a magnitude at least that large. If it does not then there has to be an error somewhere.

One might also have caught that error by tracking units more carefully. If you take the square root of a quantity in Newtons then the result will have units of $\sqrt{N}$ which is not right for a force.

 

[Eric Bretschneider]

For "extra credit" I would consider a hypothetical orbit and run some calculations from there. Since orbital velocity depends on where and object is along the orbit it might play a significant role. The velocity will change without the use of thrust (economic aspect). Just as you omitted units early in the thread, whoever is asking the question has omitted information as well.

 

[Howard Fox]

[Transcribing to LaTeX again]

The first error is on that line. What happened to the squaring that was supposed to be done on the x and y components?

Edit: Sanity checks are a useful tool. As I alluded to previously, if a force has an x component of 1.2 x 10⁸ then it should have a magnitude at least that large. If it does not then there has to be an error somewhere.

Oh wow, I made the squaring magically disappear! Thank you for spotting my distraction. So the magnitude of the resultant vector would be 1.3x10^8 N

 

[Howard Fox]

[Transcribing to LaTeX again]

The first error is on that line. What happened to the squaring that was supposed to be done on the x and y components?

Edit: Sanity checks are a useful tool. As I alluded to previously, if a force has an x component of 1.2 x 10⁸ then it should have a magnitude at least that large. If it does not then there has to be an error somewhere.

One might also have caught that error by tracking units more carefully. If you take the square root of a quantity in Newtons then the result will have units of $\sqrt{N}$ which is not right for a force.

So, I corrected the error in the acceleration calculation as well, the Y component should have been 6.0x10^-6 m/s^2, instead 6.0x10^-5, as I had previously written. So now multiplying the magnitude of the acceleration and the mass of the asteroid does give me the magnitude of the Force. Initial problem solved :)
Do you have an idea on how to proceed? Thank you very much.

 

[Howard Fox]

For "extra credit" I would consider a hypothetical orbit and run some calculations from there. Since orbital velocity depends on where and object is along the orbit it might play a significant role. The velocity will change without the use of thrust (economic aspect). Just as you omitted units early in the thread, whoever is asking the question has omitted information as well.

Yeah possibly one piece of information that I should have included is that the resultant force calculated acts in the same direction as the asteroid's motion. Other than this all the information has been included I think. Thank you

 

[gneill]

Yeah possibly one piece of information that I should have included is that the resultant force calculated acts in the same direction as the asteroid's motion. Other than this all the information has been included I think. Thank you

It's always best to post the full problem as given, and to show your work in detail. Sometimes it's a matter of missing or misinterpreting some aspect in the problem statement that catches one out.

 

[Howard Fox]

It's always best to post the full problem as given, and to show your work in detail. Sometimes it's a matter of missing or misinterpreting some aspect in the problem statement that catches one out.

Okay, thank you for the heads up. Here is a screenshot of the problem:

And here is the part I asked help for

:

Part A was just asking me to find the resultant magnitude of the force, which as outlined in earlier messages, is 1.3x10^8 N.

 

[gneill]

Posting that and the part (a) too would likely have saved everyone a lot of time

 

[Howard Fox]

Posting that and the part (a) too would likely have saved everyone a lot of time

Part A is just asking me to find the resultant magnitude of the force, which as outlined in earlier messages, is 1.3x10^8 N. I will edit my previous message to include this information.
Cheers.

 

[gneill]

Part A is just asking me to find the resultant magnitude of the force, which as outlined in earlier messages, is 1.3x10^8 N. I will edit my previous message to include this information.
Cheers.

No need, the thread has plaid out. Making changes to previous posts that have been part of the conversation will only confuse future readers.

 

[jbriggs444]

So, I corrected the error in the acceleration calculation as well, the Y component should have been 6.0x10^-6 m/s^2, instead 6.0x10^-5, as I had previously written. So now multiplying the magnitude of the acceleration and the mass of the asteroid does give me the magnitude of the Force. Initial problem solved :)

So you had calculated a force magnitude of 1.3 x 10⁸ N. With the asteroid mass at 7.5 x 10¹² kg, you have determined that the resulting acceleration has a magnitude of 6.0 x 10^-5 m/s²

Can you show us your work for that? What equation did you use? What calculation did you make?

 

[Howard Fox]

So you had calculated a force magnitude of 1.3 x 10⁸ N. With the asteroid mass at 7.5 x 10¹² kg, you have determined that the resulting acceleration has a magnitude of 6.0 x 10^-5 m/s²

Can you show us your work for that? What equation did you use? What calculation did you make?

I have used this to calculate the magnitude of the acceleration: The resulting acceleration being 1.7x10^-5 m/s^2

 

[jbriggs444]

I have used this to calculate the magnitude of the acceleration: The resulting acceleration being 1.7x10^-5 m/s^2

OK. I had managed to confuse myself into taking a different number as your result.

Note that you did not have to apply the Pythagorean theorem again. You already had the magnitude of the force available. Divide by the mass and you have the magnitude of the acceleration.

Now the only question remaining is how long it takes with an acceleration of 1.7 x 10^-5 meters/second² to achieve a delta V of 1 meter/second.

 

[Howard Fox]

OK. I had managed to confuse myself into taking a different number as your result.

Note that you did not have to apply the Pythagorean theorem again. You already had the magnitude of the force available. Divide by the mass and you have the magnitude of the acceleration.

Now the only question remaining is how long it takes with an acceleration of 1.7 x 10^-5 meters/second² to achieve a delta V of 1 meter/second.

Yeah, that's the question. I am still not sure on how to approach it. Should I try and find a value for the initial or final velocity?

 

[jbriggs444]

Yeah, that's the question. I am still not sure on how to approach it. Should I try and find a value for the initial or final velocity?

It is way easier than that. Let's try some algebra.

$a=\frac{\Delta v}{\Delta t}$

You've just finished calculating a.
You've been given $\Delta v$
You are trying to calculate $\Delta t$

 

[Howard Fox]

It is way easier than that. Let's try some algebra.

$a=\frac{\Delta v}{\Delta t}$

You've just finished calculating a.
You've been given $\Delta v$
You are trying to calculate $\Delta t$

Wow, perfect, it really was easy!
You multiply by change in t on both times and divide by acceleration on both sides. You solve for t and convert the resulting number from seconds to hours and you do get approximately 16 hours! Thank you so much, your help has been prodigious!