Help with Satellites and Grav.

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The discussion focuses on solving a physics problem involving a projectile fired from a rail gun on planet T, requiring the determination of its initial speed (vo) and launch angle (θ). Key concepts include the application of conservation of energy and angular momentum, as the projectile reaches a maximum height of ¼ R with a known final speed of 75.0 m/s. Participants emphasize the importance of considering both vertical and horizontal components of motion, while also noting that the projectile does not follow a stable circular orbit. The conversation highlights the need to relate initial and final velocities through energy conservation and angular momentum principles. Overall, the problem requires a careful analysis of gravitational forces and projectile motion dynamics.
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Homework Statement


The planet T is a planet of mass M and radius R, and very thin atmosphere (no air resistance). A rail gun has been mounted on the surface of T at the North Pole. A projectile of mass m is fired form the rail gun with an unknown speed vo at an unknown angle θ with respect to the local horizontal. The projectile is observed to rise to a maximum height above the surface of ¼ R. At this maximum height the projectile has a speed of 75.0 m/s. If M=1.5 x 10^20 (such that GM=1.0 x 1010 Nm2/kg) and R=200km, find vo in m/s. and the launch angle θ.


Homework Equations


F=ma
F=(GMm)\R^2
ma=(GMm)\R^2
a=(v^2)\r


The Attempt at a Solution


I know that F=ma, and Gravitational F=(GMm)\R^2 so ma=(GMm)\R^2 , a in rotational is a=(v^2)\r , so I would set m((v^2)\R)=(GMm)\R^2 . Yet is v in this equation the one they want? In finding the θ once I found the v would there be components in the x and y direction? Can I find it that way? Thank you very much for the help.
 
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The problem doesn't say that the projectile is an a stable [circular] orbit at this point, so that means that it may not obey circular motion. I would say that you would be expected to use conservation of energy here. It would be useful to note that at maximum height vy=0.
 
Can see what you are saying about consevation of energy, but how will that help me with finding vo
 
Hootenanny said:
Well, in both the x and y directions you know the final velocities and you know the distance traveled in the y direction. So consider the change in energy between R = 200km and the projectile's final height. This should allow you to find the vertical component of V0. You also know the final speed, from that you can calculate the horizontal component of the final velocity which of course is equal to V0 since drag is negligible. All this of course is ignoring the rotation of the earth.

Hootenanny, I don't think that it is possible here to work with a y axis. By the time the projectile has reached the maximum height, it will be at a point that may be so far from the original point that the y-axis has changed quite a bit compared to the initial y axis, if you know what I mean.

I think one must use conservation of energy (with - \frac{G m M }{ r} of course ) but one must also use angular momentum to answer this question.

Just a thought.
 
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I think your right nrqed, I've just reread the OP and with a maximum of height of R/4, my method isn't valid. Good catch, hopefully the OP will read this before he/she starts work on it. My apologies Newtonistheman :redface:, time to get some sleep I think...:zzz:
 
How should I go about finding the angle?
 
Try this way

1. Conservation of energy: potential+kinetic energy conserved

===> relation of v0 and vmax==>v0

2. Conservation of angular momentum (due to central force)

===> relation of v0cos(theta) and vmax==>theta
 
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