Rate of change in momentum of earth with direction
-2
The Earth moves around the Sun in a circle of radius 1.5 x 10^11 m at a speed of 3.0 x 10^4 m/s.The mass of the Earth is 6.0 x 10^24 kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
asked 15 hours ago
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-2
The Earth moves around the Sun in a circle of radius 1.5 x 10^11 m at a speed of 3.0 x 10^4 m/s.The mass of the Earth is 6.0 x 10^24 kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
asked 15 hours ago
rhynard patron
2
New contributor
rhynard patron is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
15 hours ago
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
14 hours 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
13 hours 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
5 hours ago
add a comment |
-2
-2
-2
The Earth moves around the Sun in a circle of radius 1.5 x 10^11 m at a speed of 3.0 x 10^4 m/s.The mass of the Earth is 6.0 x 10^24 kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
asked 15 hours ago
rhynard patron
2
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 1.5 x 10^11 m at a speed of 3.0 x 10^4 m/s.The mass of the Earth is 6.0 x 10^24 kg. Calculate the magnitude of the rate of change of the momentum of the Earth from these data.
physics
physics
asked 15 hours ago
rhynard patron
2
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 15 hours ago
rhynard patron
2
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 15 hours ago
rhynard patron
2
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 15 hours ago
rhynard patron
2
asked 15 hours ago
rhynard patron
2
2
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.
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
15 hours ago
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
14 hours 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
13 hours 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
5 hours ago
add a comment |
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
15 hours ago
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
14 hours 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
13 hours 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
5 hours ago
3
3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
15 hours ago
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
15 hours ago
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
14 hours ago
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
14 hours 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
13 hours 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
13 hours 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
5 hours 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
5 hours ago
add a comment |
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3
Hint: $$a=frac{v^2}r,quad F=ma$$Does this help?
– John Doe
15 hours ago
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
14 hours 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
13 hours 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
5 hours ago