Help with Satellites and Grav.

[newtonistheman]

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.

 

[Hootenanny]

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 v_y=0.

 

[newtonistheman]

Can see what you are saying about consevation of energy, but how will that help me with finding vo

 

[nrqed]

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 [itex] - \frac{G m M }{ r} [/itex] of course ) but one must also use angular momentum to answer this question.

Just a thought.

 

[Hootenanny]

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 , time to get some sleep I think...:zzz:

 

[newtonistheman]

How should I go about finding the angle?

 

[Weimin]

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