Modifying Newton's Kinematic Equations for High Altitude

  • #1
59
0
I recently watched a video on youtube where a guy fires a 50 cal. rifle straight up into the air and measures the time of flight at about 100s. It got me thinking about what altitude the round reached. So I used the kinematics equations and obtained a value of 12.25km. Amazing, but I wondered whether the varying magnitude of acceleration due to gravity at increasing altitude would have a significant effect. I looked up anything I could find that hints at kinematic equations for varying gravitational acceleration and had no luck.
All I could think of was to write the acceleration as a function of height and solve the new equation for height. Is this allowed? Below is what I put together. Obviously ignoring air resistance for now. Any comments/suggestions would be helpful. Am I on the right track?

Starting with $$s=vt-\frac{1}{2}at^{2}$$
Where ##s## is the height, ##v##, final velocity, ##t##, time of flight and ##a## the acceleration
Then changing ##a=-g## to $$a=-\frac{Gm_{\oplus }}{\left ( r_{\oplus }+s \right )^{2}}$$
Where ##m_{\oplus }## and ##r_{\oplus }## are respectively the mass and radius of the Earth
At maximum height ##v=0##, so the equation becomes $$s=\frac{Gm_{\oplus }t^{2}}{2\left ( r_{\oplus }+s \right )^{2}}$$
Rearranging gives a cubic in ##s##
$$2s^{3}+4r_{\oplus }s^{2}+2r_{\oplus }^{2}s-Gm_{\oplus }t^{2}=0$$
Then subbing in values of the constants ##r_{\oplus }=6.378\times 10^{6}\mathrm{m}##, ##m_{\oplus }=59.736\times 10^{23}\mathrm{kg}##, ##G=6.673\times 10^{-11}\mathrm{m^{3}kg^{-1}s^{-2}}## and ##t=50\mathrm{s}## (assuming symmetrical time of flight about the max height).
$$2s^{3}+2.5512\times 10^{7}s^{2}+8.1357768\times 10^{13}s-3.98618328\times 10^{18}=0$$
Solving (calculator) gives ##s=12.202\mathrm{km}## plus 2 complex roots. The real value is less than when ignoring variable gravitational field strength. Error or explanation? It doesn't make intuitive sense to me at all. It should be more.
 

Answers and Replies

  • #2
Bandersnatch
Science Advisor
2,967
1,934
It doesn't make intuitive sense to me at all. It should be more.
Remember that you're not comparing how high the same rifle would put a bullet in different gravity fields. You're comparing what is the height at which the bullet stops in different gravity fields, providing the deceleration from whatever initial velocity takes the same time t (or, equivalently, from what height must a bullet be dropped to reach the ground in the given time). By necessity, you'd need to use two different rifles (different ##V_0##)
E.g. imagine a very tiny asteroid, with a very tiny gravity field. Would you need to drop the bullet from very high or very low in order for it to hit the ground in 100s?

(note: I have not checked your calculations)
 
  • #3
20,844
4,542
Your starting equation is valid only for constant acceleration, which isn't the case in your problem. Are you familiar with the concepts of kinetic energy and potential energy? What do you calculate for the gravitational acceleration at 12.5 km? Have you had any calculus?
 
  • #4
59
0
I understand that the kinematic equations work for constant acceleration but they're derived from first principles by integrating with respect to time. I don't see how replacing ##a## for a function of height would have any effect on the derivations. But should we rederive suitable equations by rewriting ##a=\frac{dv}{dt}## so that we introduce the displacement variable? How would I do that? Use the chain rule so ##a=\frac{dv}{dr}\cdot \frac{dr}{dt}## and integrate with respect to ##r##? Bit flummoxed by this.
Also, I don't see how making energy considerations makes any difference. We still have a variable ##g## in ##mgh## and I also don't have a value for ##v## to use the kinetic energy. That's why I gravitated to kinematics because it could be solved just with the time of flight.
I understand that I'm not finding the height of a bullet in different constant gravitational fields. I thought having the acceleration as a function of height would solve that problem.
Thanks for your input
 
  • #5
20,844
4,542
I understand that the kinematic equations work for constant acceleration but they're derived from first principles by integrating with respect to time. I don't see how replacing ##a## for a function of height would have any effect on the derivations.
Really. If a is varying with height, it is likewise varying with time during the flight (a = a(t)). Do you still think it would have no effect on the derivation?
But should we rederive suitable equations by rewriting ##a=\frac{dv}{dt}## so that we introduce the displacement variable? How would I do that? Use the chain rule so ##a=\frac{dv}{dr}\cdot \frac{dr}{dt}## and integrate with respect to ##r##? Bit flummoxed by this.
This is a good starting point, and leads to the same results as the energy approach. So you have $$v\frac{dv}{dr}=-g\frac{r_e^2}{r^2}$$where ##r_e## is the radius of the earth at ground level (6371 km) and g is the acceleration of gravity at ground level (9.81 m/sec^2). The first step here is to find the velocity v as a function of r. Do you know how to integrate the equation to find that?

Incidentally, the acceleration of gravity at 12.5 km is $$a=9.81\left(\frac{6371}{6383.5}\right)^2=9.77\ m/s^2$$In your judgment, is the difference even worth worrying about? If you want to use your constant acceleration relationship, a good approximation would be to use the average acceleration of 9.79 m/s^2. That should be accurate enough.

I should also mention that, unlike the effect of gravitational acceleration variation with altitude, air resistance will have a substantial effect on the ultimate altitude that the bullet reaches.
 
Last edited:
  • Like
Likes nasu
  • #6
jbriggs444
Science Advisor
Homework Helper
9,567
4,233
The first chart of velocity versus distance downrange that I found online had a muzzle velocity of 2820 fps and a velocity of 2630 fps at 100 yards downrange.

That is a decrease of 190 feet per second over a travel distance of 300 feet. At an average velocity somewhere between 2630 and 2820 fps, that travel distance would have taken something just over a tenth of a second (somewhere between .114 and .106 seconds, call it 0.11 seconds).

190 feet per second in 0.11 seconds works out to an acceleration of 1727 feet per sec2. That's in the neighborhood of fifty gees.

A decision to ignore air resistance may not be wise if an accurate result is desired.
 
  • Like
Likes nasu

Related Threads on Modifying Newton's Kinematic Equations for High Altitude

  • Last Post
Replies
18
Views
1K
Replies
10
Views
1K
Replies
4
Views
7K
  • Last Post
Replies
2
Views
982
  • Last Post
Replies
10
Views
9K
Replies
1
Views
1K
Replies
1
Views
1K
  • Last Post
Replies
2
Views
3K
Top