Playing baseball

[BMAD]

A baseball player hit a ball high into the air. It took a perfectly linear path and came straight back down through the point where the man made contact after 8 seconds. Of course the ball only appeared to travel Linearly to us bystanders.  Since the earth was spinning throughout the 8 second flight of the ball,  describe the true flight the ball took during this period assuming  earth's core is (0,0,0),  make any other assumptions you need except leave the height of the ball unknown. 

[bonanova]

First thoughts:

Spoiler

1. According to an earth observer a ball goes up and down along a straight line, spending 8 seconds in the air. The problem is to describe the ball's "true" flight.
    
2. Let's examine the conditions of the problem. An earth observer sees the ball return to the point of contact after an 8-second flight. She discerns the path to follow, without the slightest deviation, a straight line. Consider the line defined by the center of the earth and the point of contact. This is the line of action for earth's gravity. If gravity is to return the ball to the point of contact, the straight line seen by the earth observer must align with gravity. That is, the ball was observed to change its radial coordinate only -- first ascending then descending, never deviating to the north, south, east or west. We conclude from the conditions of the problem that the longitude and latitude of the ball do not change during flight.
    
3. Now let's define the coordinate system to be used to describe the trajectory of the ball. Centering it on the earth's core is a beginning. Let us also put the north pole at (0, 0, R) where R (call it 8000 miles) is the earth's radius. Finally let's say that the (x, y) plane does not rotate with the earth. That is, a point fixed to the earth's equator traces a circle of radius R every 24 hours. In the plane, that point has an instantaneous (due-eastward) speed of (16000pi/24) R ~ 2094.4 mph. Points fixed to the earth at a (say) north latitude of equator = 0 < a < 90 = north pole trace smaller circles in the same time interval and have an instantaneous (due-eastward) ~ 2094.4 cos( a ) mph. So the ball's path in ( x , y, z) space depends on the latitude of the point of contact.
    
4. Let's take the easiest case first. The ball was hit at the north pole. Here the earth's rotation does not impart an initial eastward speed. There is no "east" at the north pole, only south. So the ball is rotating (CCW viewed from above) at a rate of 1 rotation per 24 hours. The ball is then hit straight upward (along the positive z coordinate), spends 8 seconds in the air, and returns, straight downward. Just as the earth observer, and us bystanders, saw it happen. End of story. I could calculate how high it went, but that is supposed to be left as an unknown.
    
5. Next let's consider what happens at the equator. When the ball is struck, it's moving eastward at ~ 2094.4 mph, as is the batter and his bat. And, according to our earth observer, it moves straight upward, staying in the ( x , y , 0) plane, just at a greater distance from the origin (0, 0, 0). Now we run into a problem. We must choose what to assume happens next. Either we believe the conditions of the problem or we believe the laws of physics.
   
   (a) Let's choose to believe the laws of physics. They require the angular momentum of the ball (as it moves in orbit around the center of the earth) to remain constant. Angular momentum is just the linear momentum (mv) of the ball multiplied by its distance (r) from the center of the earth. This means that vr remains constant. As r increases, v correspondingly decreases. But v of the batter and his bat does not decrease. So, during the flight of the ball, the batter and his bat travel eastward slightly farther than the ball does, and the ball does not return to the point of contact. Oops. According to the laws of physics, this experiment could not have been conducted on the equator.
   
   (I will admit to one caveat here: It could return to the bat, but its path would not appear to be straight: That is, if the ball were hit, with the eastward velocity needed to make up what it loses to maintain angular momentum, it could then return to the point on the bat where it was hit. And the OP permits an initial direction that is not straight up. But in this case the observed path would be curved slightly. It would appear initially to go slightly to the east, but decreasingly so until it reaches its apex, then it would retrace that curved path to the bat.)
   
   (b) Let's choose to ignore physics. That makes things much easier. We can say the ball travels to a height (say) of 200 feet instantaneously, accelerates eastward by the ratio (R+200 feet)/ R , hovers for 8 seconds, and then returns instantaneously to the bat. That provides a sketch of the ball's true path: r increases by 200 feet for 8 seconds then decreases by the same amount.
    
6. Finally, consider what happens (say) at North latitude a = 45^o. Doesn't matter. Now things get worse.
   
   (a) If we invoke the laws of physics, not only will the ball slow from its initial eastward direction to conserve angular momentum, it will also begin to travel southward. If this is difficult to visualize, consider that it is (briefly) in an elliptical orbit that has the earth's center as one focus. That focus lies to the south of the point of launch. So the plane of the ball's orbit is inclined with respect to the plane of constant latitude, in which the batter and his bat, and all of us observers, remain. Ooops. According to the laws of physics this experiment could not have been conducted at 45^o latitude (or at any latitude other than 90^o.)
   
   (b) Let's choose to ignore physics. Then, as above, we can simply say the ball travels 200 feet above the bat, maintains latitude and longitude for 8 seconds, then returns to the bat.
    
7. There is one other choice we could make. We could say the ball follows a path, governed by the laws of physics, that would be observed if the earth were not rotating. That the ball would ascend in height according to a parabola in time, reach an apex, then return to the bat after an identical time interval. Ignoring the fact that this path cannot occur if the earth is rotating and the laws of physics are obeyed, that is, if the ball acted up only by the bat (initially) and the acceleration of gravity (thereafter,) what would that path look like in an external coordinate system?
   
   (a) On the north pole it would be the same as above.
   
   (b) On the equator, it would be a path in the ( x , y , 0) plane, approximating a parabola, but with a broader apex, due to its artificially increased eastward speed. That is, in radial coordinates, the parametric equations would be r = normal parabola in time; theta = k t, where k is the angular speed of the earth.
   
   (c) At other latitudes, the path would be traced on a cone whose axis is the z axis and whose vertex is (0, 0, 0). In spherical coordinates, parametrically, r = normal parabola in time, theta = k t where k is the angular speed of the earth, and phi = a, the latitude where the ball was hit.

 

[BMAD]

On 5/2/2016 at 1:55 AM, bonanova said:

First thoughts:

  Reveal hidden contents

1. According to an earth observer a ball goes up and down along a straight line, spending 8 seconds in the air. The problem is to describe the ball's "true" flight.
    
2. Let's examine the conditions of the problem. An earth observer sees the ball return to the point of contact after an 8-second flight. She discerns the path to follow, without the slightest deviation, a straight line. Consider the line defined by the center of the earth and the point of contact. This is the line of action for earth's gravity. If gravity is to return the ball to the point of contact, the straight line seen by the earth observer must align with gravity. That is, the ball was observed to change its radial coordinate only -- first ascending then descending, never deviating to the north, south, east or west. We conclude from the conditions of the problem that the longitude and latitude of the ball do not change during flight.
    
3. Now let's define the coordinate system to be used to describe the trajectory of the ball. Centering it on the earth's core is a beginning. Let us also put the north pole at (0, 0, R) where R (call it 8000 miles) is the earth's radius. Finally let's say that the (x, y) plane does not rotate with the earth. That is, a point fixed to the earth's equator traces a circle of radius R every 24 hours. In the plane, that point has an instantaneous (due-eastward) speed of (16000pi/24) R ~ 2094.4 mph. Points fixed to the earth at a (say) north latitude of equator = 0 < a < 90 = north pole trace smaller circles in the same time interval and have an instantaneous (due-eastward) ~ 2094.4 cos( a ) mph. So the ball's path in ( x , y, z) space depends on the latitude of the point of contact.
    
4. Let's take the easiest case first. The ball was hit at the north pole. Here the earth's rotation does not impart an initial eastward speed. There is no "east" at the north pole, only south. So the ball is rotating (CCW viewed from above) at a rate of 1 rotation per 24 hours. The ball is then hit straight upward (along the positive z coordinate), spends 8 seconds in the air, and returns, straight downward. Just as the earth observer, and us bystanders, saw it happen. End of story. I could calculate how high it went, but that is supposed to be left as an unknown.
    
5. Next let's consider what happens at the equator. When the ball is struck, it's moving eastward at ~ 2094.4 mph, as is the batter and his bat. And, according to our earth observer, it moves straight upward, staying in the ( x , y , 0) plane, just at a greater distance from the origin (0, 0, 0). Now we run into a problem. We must choose what to assume happens next. Either we believe the conditions of the problem or we believe the laws of physics.
   
   (a) Let's choose to believe the laws of physics. They require the angular momentum of the ball (as it moves in orbit around the center of the earth) to remain constant. Angular momentum is just the linear momentum (mv) of the ball multiplied by its distance (r) from the center of the earth. This means that vr remains constant. As r increases, v correspondingly decreases. But v of the batter and his bat does not decrease. So, during the flight of the ball, the batter and his bat travel eastward slightly farther than the ball does, and the ball does not return to the point of contact. Oops. According to the laws of physics, this experiment could not have been conducted on the equator.
   
   (I will admit to one caveat here: It could return to the bat, but its path would not appear to be straight: That is, if the ball were hit, with the eastward velocity needed to make up what it loses to maintain angular momentum, it could then return to the point on the bat where it was hit. And the OP permits an initial direction that is not straight up. But in this case the observed path would be curved slightly. It would appear initially to go slightly to the east, but decreasingly so until it reaches its apex, then it would retrace that curved path to the bat.)
   
   (b) Let's choose to ignore physics. That makes things much easier. We can say the ball travels to a height (say) of 200 feet instantaneously, accelerates eastward by the ratio (R+200 feet)/ R , hovers for 8 seconds, and then returns instantaneously to the bat. That provides a sketch of the ball's true path: r increases by 200 feet for 8 seconds then decreases by the same amount.
    
6. Finally, consider what happens (say) at North latitude a = 45^o. Doesn't matter. Now things get worse.
   
   (a) If we invoke the laws of physics, not only will the ball slow from its initial eastward direction to conserve angular momentum, it will also begin to travel southward. If this is difficult to visualize, consider that it is (briefly) in an elliptical orbit that has the earth's center as one focus. That focus lies to the south of the point of launch. So the plane of the ball's orbit is inclined with respect to the plane of constant latitude, in which the batter and his bat, and all of us observers, remain. Ooops. According to the laws of physics this experiment could not have been conducted at 45^o latitude (or at any latitude other than 90^o.)
   
   (b) Let's choose to ignore physics. Then, as above, we can simply say the ball travels 200 feet above the bat, maintains latitude and longitude for 8 seconds, then returns to the bat.
    
7. There is one other choice we could make. We could say the ball follows a path, governed by the laws of physics, that would be observed if the earth were not rotating. That the ball would ascend in height according to a parabola in time, reach an apex, then return to the bat after an identical time interval. Ignoring the fact that this path cannot occur if the earth is rotating and the laws of physics are obeyed, that is, if the ball acted up only by the bat (initially) and the acceleration of gravity (thereafter,) what would that path look like in an external coordinate system?
   
   (a) On the north pole it would be the same as above.
   
   (b) On the equator, it would be a path in the ( x , y , 0) plane, approximating a parabola, but with a broader apex, due to its artificially increased eastward speed. That is, in radial coordinates, the parametric equations would be r = normal parabola in time; theta = k t, where k is the angular speed of the earth.
   
   (c) At other latitudes, the path would be traced on a cone whose axis is the z axis and whose vertex is (0, 0, 0). In spherical coordinates, parametrically, r = normal parabola in time, theta = k t where k is the angular speed of the earth, and phi = a, the latitude where the ball was hit.

 

Am I understanding your claim correctly? Are you stating that it is impossible for a ball to give the appearance to a bystander of traveling straight up and down over the course of 8 seconds? 

[bonanova]

I think only at the North Pole this could happen. Or on a non-spinning globe. See Coriolis Force.

In real life, a ball hit up in the air by a bat has great spin, and follows a curved path. In an idealized sense, with no wind resistance, and in an inertial frame of reference, a ball could go straight up and straight down. But the thing that makes this an interesting question (earth's rotation) also prevents that in general from happening. Rotation is acceleration, and it introduces forces (centrifugal and Coriolis) in addition to gravity. Over short distances and small times, their effect is small.

[the_count]

I think Bonanova's assumptions make sense..

It seems hard to provide definition for the 'true' path of the ball. Even the centre of the Earth is an arbitrary frame, we could have used the Galactic plane?

[Philip]

If without air resistance and not on North or south pole, parabolic, because of the initial velocity given by the spin of the earth while still on hand, similar to throwing the ball with some initial velocity, and actually seeing the parabolic projectile shape. if on North or south pole, linear. If taking into account the speed of the motion of earth around the solar system (revolution), the rotation of earth would be negligible, thus it is near elliptical (same with galactic reference of sun orbiting galaxy).

 

[BMAD]

On 5/5/2016 at 1:06 PM, the_count said:

I think Bonanova's assumptions make sense..

It seems hard to provide definition for the 'true' path of the ball. Even the centre of the Earth is an arbitrary frame, we could have used the Galactic plane?

This is what I had in mind