Here's the question you clicked on:
heena
why the escape velocity is independent of mass of projecting object and angle of projection?
@Callisto @Mani_Jha @apoorvk @Diyadiya @shivam_bhalla
See \[{GM_e M_o \over R_e +h}= {1 \over 2} m_o v^2\]
M_e --> Mass of earth R_e -->Radius of Eart M_o-->Mass of projectile v-->escape velocity
The above equation is by energy conservation
buddhi lagao. Energy concept. the potential barrier has to be crossed. apply shivam's equation above.
So M_o cancels and as @apoorvk said, buddhi lagao :P
lol.. anyways i need a theoretical ans if u guys see the formulae there is a presence of mass isntd ?? which show it dependent on mass right?
R---> Radius of earth h--> height of projectile from earth
@heena , "Open your eyes" It shows dependence on "MASS OF THE EARTH"
Okay you mean you want a 'feel'? Hmm. Okay.
shiv what is buddhi lagao ?
"Use your brains " :P
mean use ur brain grrrrrrr.........
buddhi lagao ---> translates to --> apply brains
@heena , aur kuch (anything else ) ??
ok means it is depends on mass but mass of earth right :) thnk you all of u meri khinchai krne k liye :P popat kr diya mera hur. :D
Main bhi popat bane ke bad sikha ye concept. So don't worry :D
ye concept *sikha
janti hu par guys y qn mera ni tha :P some one asked or m physc m apna dimag ni lagati sirf fimrulae se mtlb rakhti hu :P baki to aplog ho na :D :D
but one thing surely i m not going to forgot this concept ever :P
LOL Remember only the vital Formulas and derice the rest using conservation of 1) Energy 2)linear momentum 3)Angular momentum
bhalla u free?? mje isme se ek bhi derivation ni ati frnkly sayin i m so dumb plz help me if u have time tell me how we do derivation :?
Sorry not *free exactly but if you want help in any specific derivation, then I can help you
actaully in 11 my teacher dosent teach v.well thi topic i mean conservation of laws dats y i m facing problems.. :(
Go through the basics of work , gravitational force, kinetic energy, linear momentum, angular momentum, Then for any derivation,you have to apply only these 2 or 3 laws conservation of 1) Energy 2)linear momentum 3)Angular momentum If you have any doubt, you can post a question here and tag me. I will try t help you with that derivation
ok thnk you guys for being here i have to go to study and yea i ll shivam and one qn @shivam yeterday was ur bday?
dw m yaha sbko tang krne k liye hi hu :D :D and belated bday sorry unable to find u yeterday... many many returns of the belated day... and orry once again for being late to wish u...
am asking why the angle is independ of the escape velocity?
LOL, Ofcourse @swalah , put an angle of projection will increase your escape velocity and escape velocity is the "minimum" velocity required to escape the gravity of earth
LOL, Ofcourse @swalah , "putting " an angle of projection will increase your escape velocity and escape velocity is the "minimum" velocity required to escape the gravity of earth
thanks then what is the angle of that minimum velocity? @shivam_bhalla
Ofcourse, 90 degrees
you assume space is spherical yeah?
The gravitational force of the the earth acts spherically as a function of r.
from behalf of @Mani_Jha It doesn't matter whether you throw a bus or a ball into space. Both of them need the same speed 11.6 km/sec to do that(escape velocity of earth). A possible reason of that is that the gravitational field(g) is same for all objects on earth. All objects are in the same field(gravitational) and so require the same velocity to escape the earth. Also, whatever the object be, it has to travel the same distance from the ground to space. Although, it is more difficult to give a bus a speed of 11.6 km/sec than to a ball. Whether you throw it vertically upwards or at an angle, it doesn't matter for the speed. This is because, as shivam_bhalla said, the earth is spherically symmetrical(roughly). Look above at the sky. There is space everywhere at the same distance from you. So, from wherever you throw it, if it has a speed of 11.6 km/sec, it will reach space. , as I think that @shivam_bhalla might have been mistaken.
@heena , You are right. :D I was half sleepy . thanks for correcting me. :)
no give thanks to @Mani_Jha as i mention in behalf of him i m posting the answer..
he is having trouble in posting answer.. :(
Thanks @Mani_Jha . and @heena , I was just about to ask about that