If something MASSIVE were to be near Earth

[DrPizza]

The mathematics is pretty simple to figure out, assuming the motion is relatively straight. From where you're standing on the ground, mount two green lasers pointing up into the sky, each at some angle from you. It really doesn't matter what the angle is for the following point, but let's use a 45 degree angle. And, for simplicity's sake, let's say the object is moving parallel to a tangent line to the earth at the point you're standing and that those lasers intersect that line. In this case, the lasers and motion of the object form the 3 sides of a triangle. If the object is 100 feet above the Earth, and moves from one laser beam and across your field of view to the other in t seconds, then at 200 feet above the Earth, all you've done is created a similar triangle (dilation by a factor of 2). It must now travel a path twice as long; but if moving at the same speed relative to the Earth, it'll take twice as much time, so will appear to be moving slower - 1/2 the speed. At 300 feet, it'll look like it's moving 1/3 the speed, and so on. So, comparing 100 feet above the Earth to 1000 feet above the Earth, it'll only be moving across your field of view at 10% the original rate.

One caveat - as far as the number of degrees per second, that will not be a constant rate if the object is following a linear path at a constant speed. A mental exercise, if the OP is capable, is to imagine how fast your hand is moving if you're standing on the side of a very straight road and pointing at a moving car. If you keep your finger pointed at the car, and track it with your finger from the time it's 1/2 a mile to the left of you until the time when it's 1/2 a mile to the right of you, you'll realize that as it passes in front of you, your finger is moving a lot faster than when it's farther away from you.

This does point out though, since the difference between the average degrees per second across the field of view varies so much from 100 feet to 1000 feet, there's no real way to answer the OP very well, since even 100,000,000 feet is considered to be *near* the Earth for massive objects such as he mentioned.

 

[JulesMaximus]

I've seen the ISS from the ground and it moves pretty fast across the sky. It is travelling at close to 18,000 mph and orbits the earth once every 90 minutes or so.

The ISS is about 200' x 300'

 

[JM Aggie08]

Similar to when your mother rolls over in bed?

 

[JTsyo]

The mathematics is pretty simple to figure out, assuming the motion is relatively straight. From where you're standing on the ground, mount two green lasers pointing up into the sky, each at some angle from you. It really doesn't matter what the angle is for the following point, but let's use a 45 degree angle. And, for simplicity's sake, let's say the object is moving parallel to a tangent line to the earth at the point you're standing and that those lasers intersect that line. In this case, the lasers and motion of the object form the 3 sides of a triangle. If the object is 100 feet above the Earth, and moves from one laser beam and across your field of view to the other in t seconds, then at 200 feet above the Earth, all you've done is created a similar triangle (dilation by a factor of 2). It must now travel a path twice as long; but if moving at the same speed relative to the Earth, it'll take twice as much time, so will appear to be moving slower - 1/2 the speed. At 300 feet, it'll look like it's moving 1/3 the speed, and so on. So, comparing 100 feet above the Earth to 1000 feet above the Earth, it'll only be moving across your field of view at 10% the original rate.

One caveat - as far as the number of degrees per second, that will not be a constant rate if the object is following a linear path at a constant speed. A mental exercise, if the OP is capable, is to imagine how fast your hand is moving if you're standing on the side of a very straight road and pointing at a moving car. If you keep your finger pointed at the car, and track it with your finger from the time it's 1/2 a mile to the left of you until the time when it's 1/2 a mile to the right of you, you'll realize that as it passes in front of you, your finger is moving a lot faster than when it's farther away from you.

This does point out though, since the difference between the average degrees per second across the field of view varies so much from 100 feet to 1000 feet, there's no real way to answer the OP very well, since even 100,000,000 feet is considered to be *near* the Earth for massive objects such as he mentioned.

I somewhat remember long ago, in a HS somewhere they had us doing the equation for a train passing you by at a constant linear speed in terms for degrees per second. Now I forget if it was in pre-cal or Calculus.

 

[TechBoyJK]

http://vimeo.com/93615116

Cool music video made with scenes from the movie Melancholia, which is about a foreign planet entering the solar system and getting close to earth. Basically Planet X.

They know it's coming for months but seems to speed up as it gets closer.

 

[DrPizza]

I somewhat remember long ago, in a HS somewhere they had us doing the equation for a train passing you by at a constant linear speed in terms for degrees per second. Now I forget if it was in pre-cal or Calculus.

That would likely be a related rates problem in calculus. If you were standing near straight train tracks, you can form a triangle from you, to the nearest point on the tracks (which would form a right angle) to the train. The angle theta, measured between the line from you to the tracks and from you to the train. Tan(theta) = opposite/adjacent. So, if you're 50 feet from the train, and the train is 500 feet down the tracks, tan(theta)=500/50.

You're concerned with the rate that angle theta is changing. Let's say that at that moment, the train is traveling at 60 feet per second. The distance from you to the tracks is a constant 50 feet (you're not moving backward). Since tan(theta)=opp/adj, taking the derivative with respect to time gives sec^2(theta) dtheta/dt = 1/50 * dopp/dt.

the rate of change of theta with respect to time is therefore 1/50 * cos^2(theta) * rate of change of the train. If the train is traveling toward you, it's -60 ft/second. The cos(theta) can be calculated by using simple right triangle trig & the pythagorean theorem. Since the terms on the right side of that equation aren't all constants (namely, cos^2(theta), the rate of change of theta isn't going to be a constant and will depend on where the train is.

 

[dawp]

a neutron star is massive but not all that big.