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What happens when the drag force exceeds the weight of an object falling into earth?
Calculating wind force and drag force on a falling objectWhy aren't Roche limit and the difference in gravitational acceleration the same?How to calculate the distance an accelerating object will be pulled by another object in a given amount of time?How does a small object move with constant velocity when drag force is equal to its weight?Acceleration of a ball thrown into the airPower in 1st GearDrag force equals weightWhy wouldn't the weight of an object affect the magnitude of the drag force on that object?Finding the drag force from the final velocity of a falling objectThrust needed from an engine to reach orbit
$begingroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
Thanks in advance!
gravity acceleration drag
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
Thanks in advance!
gravity acceleration drag
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
Thanks in advance!
gravity acceleration drag
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let's say a meteor is coming towards earth. It's not accelerating, but it does have an initial velocity. This meteor is shaped so it has an insane amount of drag, enough to even exceed its weight (not mass) as it gets closer. What happens? Why?
At first I thought it would just stop, but that doesn't really make sense. The forces cancel but that doesn't mean the body slows down. Would it keep going in another direction?
Thanks in advance!
gravity acceleration drag
gravity acceleration drag
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
asked 2 hours ago
Laura IglesiasLaura Iglesias
182
182
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Laura Iglesias is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
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$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
$endgroup$
add a comment |
$begingroup$
Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered 2 hours ago
AllureAllure
2,361926
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
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votes
$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
$endgroup$
add a comment |
$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
$endgroup$
add a comment |
$begingroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
$endgroup$
The acceleration (or deceleration) of an object is $a$: $$a = f/m.$$
Acceleration (or deceleration) is the rate of change of velocity. So, to find the rate of change of velocity of an object you divide all the forces acting on it, by the object's mass.
The forces acting on your hypothetical object are the object's weight and the atmospheric drag. Someone being very picky might also say that there is a relatively very small atmospheric buoyancy at work too, but it can be ignored in most cases.
Let's say the object is falling straight down. Atmospheric drag is velocity-dependent, with low drag at low speeds and much higher drag at higher speeds. Drag will cause the object to decelerate, until drag equals the object's weight. At that point the object is said to be moving at terminal velocity, and it just keeps falling at terminal velocity.
The whole scenario gets more complicated when altitude-dependent atmospheric density is taken into account, but what's described above captures the essence of an answer to your question.
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
answered 1 hour ago
S. McGrewS. McGrew
9,82921239
9,82921239
add a comment |
add a comment |
$begingroup$
Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered 2 hours ago
AllureAllure
2,361926
$endgroup$
add a comment |
$begingroup$
Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered 2 hours ago
AllureAllure
2,361926
$endgroup$
add a comment |
$begingroup$
Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered 2 hours ago
AllureAllure
2,361926
$endgroup$
Drag is proportional to the object's velocity, so as the object's speed decreases, the drag decreases as well. Therefore, the body won't start "falling upwards" because for that to happen, the velocity must reach zero, and at that point the drag also drops to zero.
In practice, the falling object slows down until the drag equals the weight, at which point it keeps moving at constant speed (so-called "terminal velocity"). In the hypothetical scenario where you have a constant force that's greater than the weight pulling on the object, then the object will indeed start falling upwards.
answered 2 hours ago
AllureAllure
2,361926
edited 1 hour ago
answered 2 hours ago
AllureAllure
2,361926
answered 2 hours ago
AllureAllure
2,361926
answered 2 hours ago
AllureAllure
2,361926
2,361926
add a comment |
add a comment |
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