Rate of change in momentum of earth with direction [on hold]
The Earth moves around the Sun in a circle of radius $R=1.5 times 10^{11}$ m at a speed of $v=3.0 times 10^4$ m/s.The mass of the Earth is $m=6.0 times 10^{24}$ kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
edited yesterday
Andrei
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asked Jan 3 at 5:10
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put on hold as off-topic by John Doe, Nosrati, Saad, Shubham Johri, T. Bongers yesterday
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The Earth moves around the Sun in a circle of radius $R=1.5 times 10^{11}$ m at a speed of $v=3.0 times 10^4$ m/s.The mass of the Earth is $m=6.0 times 10^{24}$ kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
edited yesterday
Andrei
11.3k21026
asked Jan 3 at 5:10
rhynard patron
1
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by John Doe, Nosrati, Saad, Shubham Johri, T. Bongers yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Doe, Nosrati, Saad, Shubham Johri, T. Bongers
If this question can be reworded to fit the rules in the help center, please edit the question.
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
Jan 3 at 5:15
Thanks but im trying to find the rate of change of the momentum of the earth and what I know is that its magnitude does not change but its direction does. Huhuhu
– rhynard patron
Jan 3 at 5:32
The rate of change of momentum is the radial force exerted on Earth by the sun$$frac{dmathbf p}{dt}=mfrac{dmathbf v}{dt}=mmathbf a$$$mathbf a$ is the centripetal acceleration given by $frac{|mathbf v|^2}{|mathbf r|^2}(-mathbf r)$, where $mathbf r$ is the radial vector from the sun to the Earth
– Shubham Johri
2 days ago
Yes, what Shubham said is correct. Newton's 2nd Law of motion says that the rate of change of momentum is equal to the force, so $$frac{dp}{dt}=F$$
– John Doe
2 days ago
add a comment |
The Earth moves around the Sun in a circle of radius $R=1.5 times 10^{11}$ m at a speed of $v=3.0 times 10^4$ m/s.The mass of the Earth is $m=6.0 times 10^{24}$ kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
edited yesterday
Andrei
11.3k21026
asked Jan 3 at 5:10
rhynard patron
1
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The Earth moves around the Sun in a circle of radius $R=1.5 times 10^{11}$ m at a speed of $v=3.0 times 10^4$ m/s.The mass of the Earth is $m=6.0 times 10^{24}$ kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
physics
edited yesterday
Andrei
11.3k21026
asked Jan 3 at 5:10
rhynard patron
1
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Andrei
11.3k21026
asked Jan 3 at 5:10
rhynard patron
1
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited yesterday
Andrei
11.3k21026
edited yesterday
Andrei
11.3k21026
edited yesterday
Andrei
11.3k21026
11.3k21026
asked Jan 3 at 5:10
rhynard patron
1
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Jan 3 at 5:10
rhynard patron
1
asked Jan 3 at 5:10
rhynard patron
1
1
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put on hold as off-topic by John Doe, Nosrati, Saad, Shubham Johri, T. Bongers yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Doe, Nosrati, Saad, Shubham Johri, T. Bongers
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by John Doe, Nosrati, Saad, Shubham Johri, T. Bongers yesterday
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – John Doe, Nosrati, Saad, Shubham Johri, T. Bongers
If this question can be reworded to fit the rules in the help center, please edit the question.
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
Jan 3 at 5:15
Thanks but im trying to find the rate of change of the momentum of the earth and what I know is that its magnitude does not change but its direction does. Huhuhu
– rhynard patron
Jan 3 at 5:32
The rate of change of momentum is the radial force exerted on Earth by the sun$$frac{dmathbf p}{dt}=mfrac{dmathbf v}{dt}=mmathbf a$$$mathbf a$ is the centripetal acceleration given by $frac{|mathbf v|^2}{|mathbf r|^2}(-mathbf r)$, where $mathbf r$ is the radial vector from the sun to the Earth
– Shubham Johri
2 days ago
Yes, what Shubham said is correct. Newton's 2nd Law of motion says that the rate of change of momentum is equal to the force, so $$frac{dp}{dt}=F$$
– John Doe
2 days ago
add a comment |
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
Jan 3 at 5:15
Thanks but im trying to find the rate of change of the momentum of the earth and what I know is that its magnitude does not change but its direction does. Huhuhu
– rhynard patron
Jan 3 at 5:32
The rate of change of momentum is the radial force exerted on Earth by the sun$$frac{dmathbf p}{dt}=mfrac{dmathbf v}{dt}=mmathbf a$$$mathbf a$ is the centripetal acceleration given by $frac{|mathbf v|^2}{|mathbf r|^2}(-mathbf r)$, where $mathbf r$ is the radial vector from the sun to the Earth
– Shubham Johri
2 days ago
Yes, what Shubham said is correct. Newton's 2nd Law of motion says that the rate of change of momentum is equal to the force, so $$frac{dp}{dt}=F$$
– John Doe
2 days ago
3
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
Jan 3 at 5:15
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
Jan 3 at 5:15
Thanks but im trying to find the rate of change of the momentum of the earth and what I know is that its magnitude does not change but its direction does. Huhuhu
– rhynard patron
Jan 3 at 5:32
Thanks but im trying to find the rate of change of the momentum of the earth and what I know is that its magnitude does not change but its direction does. Huhuhu
– rhynard patron
Jan 3 at 5:32
The rate of change of momentum is the radial force exerted on Earth by the sun$$frac{dmathbf p}{dt}=mfrac{dmathbf v}{dt}=mmathbf a$$$mathbf a$ is the centripetal acceleration given by $frac{|mathbf v|^2}{|mathbf r|^2}(-mathbf r)$, where $mathbf r$ is the radial vector from the sun to the Earth
– Shubham Johri
2 days ago
The rate of change of momentum is the radial force exerted on Earth by the sun$$frac{dmathbf p}{dt}=mfrac{dmathbf v}{dt}=mmathbf a$$$mathbf a$ is the centripetal acceleration given by $frac{|mathbf v|^2}{|mathbf r|^2}(-mathbf r)$, where $mathbf r$ is the radial vector from the sun to the Earth
– Shubham Johri
2 days ago
Yes, what Shubham said is correct. Newton's 2nd Law of motion says that the rate of change of momentum is equal to the force, so $$frac{dp}{dt}=F$$
– John Doe
2 days ago
Yes, what Shubham said is correct. Newton's 2nd Law of motion says that the rate of change of momentum is equal to the force, so $$frac{dp}{dt}=F$$
– John Doe
2 days ago
add a comment |
1 Answer
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Draw a diagram. Show the position of the Earth at $t=0$ then at some small interval $Delta t$ afterwards. Since the Earth is moving in a circular orbit, you can describe the position just by the angle. Note also that the momentum is always tangent to the trajectory, so the angle between the initial momentum and the momentum at $Delta t$ is the same $Delta theta$ as the difference between the positions at the two times. Now if you have an isosceles triangle with two sides equal to $p$, and angle $Delta theta$ between them, you get the length of the remaining side as $$Delta p=2psinfrac{Deltatheta}2$$
In the small angle limit (when $Deltatheta$ goes to $0$), you have $$Delta p=pDeltatheta$$ Divide by $Delta t$ and put the limit $Delta tto 0$ and you get $$frac {dp}{dt}=pfrac{dtheta}{dt}$$
To calculate the change in angle, you know that you go $2pi$ radians in one period, and to get the period, you have $$v=frac{2pi R}{T}$$
Therefore $$frac {dtheta}{dt}=frac{2pi}{T}=frac vR$$
answered yesterday
Andrei
11.3k21026
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Draw a diagram. Show the position of the Earth at $t=0$ then at some small interval $Delta t$ afterwards. Since the Earth is moving in a circular orbit, you can describe the position just by the angle. Note also that the momentum is always tangent to the trajectory, so the angle between the initial momentum and the momentum at $Delta t$ is the same $Delta theta$ as the difference between the positions at the two times. Now if you have an isosceles triangle with two sides equal to $p$, and angle $Delta theta$ between them, you get the length of the remaining side as $$Delta p=2psinfrac{Deltatheta}2$$
In the small angle limit (when $Deltatheta$ goes to $0$), you have $$Delta p=pDeltatheta$$ Divide by $Delta t$ and put the limit $Delta tto 0$ and you get $$frac {dp}{dt}=pfrac{dtheta}{dt}$$
To calculate the change in angle, you know that you go $2pi$ radians in one period, and to get the period, you have $$v=frac{2pi R}{T}$$
Therefore $$frac {dtheta}{dt}=frac{2pi}{T}=frac vR$$
answered yesterday
Andrei
11.3k21026
add a comment |
Draw a diagram. Show the position of the Earth at $t=0$ then at some small interval $Delta t$ afterwards. Since the Earth is moving in a circular orbit, you can describe the position just by the angle. Note also that the momentum is always tangent to the trajectory, so the angle between the initial momentum and the momentum at $Delta t$ is the same $Delta theta$ as the difference between the positions at the two times. Now if you have an isosceles triangle with two sides equal to $p$, and angle $Delta theta$ between them, you get the length of the remaining side as $$Delta p=2psinfrac{Deltatheta}2$$
In the small angle limit (when $Deltatheta$ goes to $0$), you have $$Delta p=pDeltatheta$$ Divide by $Delta t$ and put the limit $Delta tto 0$ and you get $$frac {dp}{dt}=pfrac{dtheta}{dt}$$
To calculate the change in angle, you know that you go $2pi$ radians in one period, and to get the period, you have $$v=frac{2pi R}{T}$$
Therefore $$frac {dtheta}{dt}=frac{2pi}{T}=frac vR$$
answered yesterday
Andrei
11.3k21026
add a comment |
Draw a diagram. Show the position of the Earth at $t=0$ then at some small interval $Delta t$ afterwards. Since the Earth is moving in a circular orbit, you can describe the position just by the angle. Note also that the momentum is always tangent to the trajectory, so the angle between the initial momentum and the momentum at $Delta t$ is the same $Delta theta$ as the difference between the positions at the two times. Now if you have an isosceles triangle with two sides equal to $p$, and angle $Delta theta$ between them, you get the length of the remaining side as $$Delta p=2psinfrac{Deltatheta}2$$
In the small angle limit (when $Deltatheta$ goes to $0$), you have $$Delta p=pDeltatheta$$ Divide by $Delta t$ and put the limit $Delta tto 0$ and you get $$frac {dp}{dt}=pfrac{dtheta}{dt}$$
To calculate the change in angle, you know that you go $2pi$ radians in one period, and to get the period, you have $$v=frac{2pi R}{T}$$
Therefore $$frac {dtheta}{dt}=frac{2pi}{T}=frac vR$$
answered yesterday
Andrei
11.3k21026
Draw a diagram. Show the position of the Earth at $t=0$ then at some small interval $Delta t$ afterwards. Since the Earth is moving in a circular orbit, you can describe the position just by the angle. Note also that the momentum is always tangent to the trajectory, so the angle between the initial momentum and the momentum at $Delta t$ is the same $Delta theta$ as the difference between the positions at the two times. Now if you have an isosceles triangle with two sides equal to $p$, and angle $Delta theta$ between them, you get the length of the remaining side as $$Delta p=2psinfrac{Deltatheta}2$$
In the small angle limit (when $Deltatheta$ goes to $0$), you have $$Delta p=pDeltatheta$$ Divide by $Delta t$ and put the limit $Delta tto 0$ and you get $$frac {dp}{dt}=pfrac{dtheta}{dt}$$
To calculate the change in angle, you know that you go $2pi$ radians in one period, and to get the period, you have $$v=frac{2pi R}{T}$$
Therefore $$frac {dtheta}{dt}=frac{2pi}{T}=frac vR$$
answered yesterday
Andrei
11.3k21026
answered yesterday
Andrei
11.3k21026
answered yesterday
Andrei
11.3k21026
answered yesterday
Andrei
11.3k21026
11.3k21026
add a comment |
add a comment |
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
Jan 3 at 5:15
Thanks but im trying to find the rate of change of the momentum of the earth and what I know is that its magnitude does not change but its direction does. Huhuhu
– rhynard patron
Jan 3 at 5:32
The rate of change of momentum is the radial force exerted on Earth by the sun$$frac{dmathbf p}{dt}=mfrac{dmathbf v}{dt}=mmathbf a$$$mathbf a$ is the centripetal acceleration given by $frac{|mathbf v|^2}{|mathbf r|^2}(-mathbf r)$, where $mathbf r$ is the radial vector from the sun to the Earth
– Shubham Johri
2 days ago
Yes, what Shubham said is correct. Newton's 2nd Law of motion says that the rate of change of momentum is equal to the force, so $$frac{dp}{dt}=F$$
– John Doe
2 days ago