Physics A Level Question, Please help. :/ Parts i and ii.
@aaronq @Abhisar
please note that, potential energy of your object ii the earth gravitational field, is: \[U(d)=\frac{ GMm }{ d+R }\] where M is mass of earth, m is mass of your object, d is the altitude of your object with respect to earth surface, and R is earth radius
oops.. G is the Gravitational constant
of course, when your object is at earth surface, its potential energy is: \[U(R)=G \frac{ Mm }{ R }\]
so the change in potential energy, is: \[U(R)-U(d+R)=G \frac{ Mm }{ R }-G \frac{ Mm }{ d+R}=...\]
please insert your numerical data @MonsterIsEnergy
Oh thank you so much ! But what do you mean by U(d) and U(R) ? I get the rest of it though ^-^ I appreciate it thanks!
Oh wait
I get it now!
U(d) is the potential energy of your object at its final position, whereas U(R) is potential energy of your object at its initial position, namely before to be launched
Aha I see, but in the question, how do we use the altitude, ....
the altitude is d
All righty thank you so much ^-^ Have a great day
please, do you know how to calculate the speed of your object, at launch
Using the kinetic energy formula?
0.5 m v^2 = m X change in potential
please note that in the earth gravitational field the total energy has to be remain constant, namely \[\frac{ GMm }{ R}=+\frac{ 1 }{ 2 }mV ^{2}=G \frac{ Mm }{ d+R }+\frac{ 1 }{ 2 }m V _{f}^{2}\] where V_f is the speed of your object at its final position. I think that V_f=0, do you agree?
oops sorry: \[G \frac{ Mm }{ R }+\frac{ 1 }{ 2 }mV ^{2}=G \frac{ Mm }{ d+R }+\frac{ 1 }{ 2 }mV _{f}^{2}\]
and V is the speed of your object at initial position, namely the launch speed
Ohhhh yep that makes a lot of sense I will go over all of that once more , thanks.
thanks!!
Thank YOU ^-^
Thank YOU!!
Awh nope haha, thank YEW! xD lol.
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