A meteorite of mass m= 2×10^4 kg is approaching head-on a planet of mass M= 4×10^29 kg and radius R= 2×10^4 km. Assume that the meteorite is initially at a very large distance from the planet where it has a speed v0= 5×10^2 km/s. Take G= 6.67×10^−11.
Determine the speed of the meteorite v (in m/s) just before it hits the surface of the planet. (The planet has no atmosphere, so we can neglect all friction before impact)
v=
The potential energy of the meteorite is approximately GMm/distance.
Add that PE, and the initial KE of the meteorite, and that is the final KE of the body, 1/2 mv^2, solve for v.
What is the distance here?
My bob that approach got me a wrong answer. :(
Yes, this method is not working
try PE=0 as its at infinity
5*10^5 m/s ???
Well, think a little... If you're at an infinity distance, your gravitational potential must be zero, as you see: -mMG/distance, if distance is infinity, it doesn't metter -mMG as somehow it is some constant. So, you have some velocity at "infinity" so you have some kinect energy, (m * v0^2)/2... So you see, conservation says that Potential_A + Kinect_A must be equal to Potential_B + Kinect_B, so at B, it will be when you hit the planet... Well, now you're not anymore at some infinity distance, so you have some potential energy, from the distance of the center of the planet, that distance is R, then you have -mMG/R, and now you have some Kinect energy, m * v1^2/2, this v1^2 is the "collision" velocity. So, all you must do now, is put v1 at evidence, then you have something like this: m*v0^2/2 = -mMG/R + m*v1^2/2; cut down those "m", you have v0^2/2 = -MG/R + v1^2/2; put v1^2 at evidence "throwing" -MG/R to the otherside, you have: v1^2/2 = v0^2/2 + MG/R; "throw" 2 to the otherside: v1^2 = v0^2 (you don't need 2 here, cause it's already divided by 2) + 2MG/R; Almost there now, now you take the square root of it: v1 = sqrt(v0^2 + 2MG/R). So, now you have it. Just remember, at infinity, potential energy is zero, and remember to convert those values. I guess you'll get something like v1 = 1.708x10^6. I hope it helps. Good luck.
The formula for the meteorite problem is:
KE(initial)=KE(final) - GMm/(Radius of Earth)
Can somebody please help with the aeroplane landing and Pendulum problems? :(
hi buddies, for the airplane:
physicsforums com showthread.php?t=720405
I've worked out the pendulum and will post the solution as soon as i wake up now i need sleep, we have still 2 days left so no worries
@fellow 801xer please tell the mass pushed by spring too
guys guys ..listen..! just keep it simple ..! v_1=sqrt((v_o)^2+2GM/R). be careful and ur units must be in m/sec for velocity and kgs for mass 1km/sec=1000m/sec. 1km/sec=1000m/sec..
anyone will mass pushed by spring?
Listen... guys, I can help with all the problems barring the airliner, pendulum and mass-spring.... please let me know if you guys need any of the other answers and if you have solutions to the problems I'm stuck on?
i just need mass spring and airliner last 2 question and the block on rotating disc
ss01... tell me the answers/formulae for pendulum???
Block on rotating disk is easy - I got option (a) :)
solve for v: mv^2/R=umg....
I need help for airliner and double well potential ( K=? T=? )
And also for sliding down the dome's 2nd part
For the pendulum just review L18 video 3. For a) you should use the conservation of mechanical energy (mv(bot)^2)/2 = (mv0^2)/2 + mgh where h is the distance of the ball from the ground calculated as h = L*(1-cos(theta)). For b) put together y component of gravitational force and centrifugal force. The other problems are just motion of the projectile where you know v0 and angle.
Troll.... what's the angle of launch for the projectile???
like, suppose the string was cut at and angle alpha=70degrees, the launch angle is 70 degs too???
It is because it goes off on a tangent. The direction of motion is perpendicular to the line of the pendulum at all times.
For v_1, you have sqrt(v_0^2 - 2*G*(L(1 - cos(alpha)))), right? For tension, you use T = Fgravitational + Fcentrifugal, where Fg = m*g*cos(alpha) and Fc = m*(v_1)^2/L. Then for max height you'll have something like this: (v_1 * sin(alpha))^2/(2*G), and for alpha you use your given angle. If something is wrong, just say it.
Block on rotating disk is easy -
solve for v: mv^2/R=umg....what's u?