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Physics 16 Online
OpenStudy (monsterisenergy):

Physics A Level Question, Please help. :/ Parts i and ii.

OpenStudy (monsterisenergy):

OpenStudy (monsterisenergy):

@aaronq @Abhisar

OpenStudy (michele_laino):

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

OpenStudy (michele_laino):

oops.. G is the Gravitational constant

OpenStudy (michele_laino):

of course, when your object is at earth surface, its potential energy is: \[U(R)=G \frac{ Mm }{ R }\]

OpenStudy (michele_laino):

so the change in potential energy, is: \[U(R)-U(d+R)=G \frac{ Mm }{ R }-G \frac{ Mm }{ d+R}=...\]

OpenStudy (michele_laino):

please insert your numerical data @MonsterIsEnergy

OpenStudy (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!

OpenStudy (monsterisenergy):

Oh wait

OpenStudy (monsterisenergy):

I get it now!

OpenStudy (michele_laino):

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

OpenStudy (monsterisenergy):

Aha I see, but in the question, how do we use the altitude, ....

OpenStudy (michele_laino):

the altitude is d

OpenStudy (monsterisenergy):

All righty thank you so much ^-^ Have a great day

OpenStudy (michele_laino):

please, do you know how to calculate the speed of your object, at launch

OpenStudy (monsterisenergy):

Using the kinetic energy formula?

OpenStudy (monsterisenergy):

0.5 m v^2 = m X change in potential

OpenStudy (michele_laino):

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?

OpenStudy (michele_laino):

oops sorry: \[G \frac{ Mm }{ R }+\frac{ 1 }{ 2 }mV ^{2}=G \frac{ Mm }{ d+R }+\frac{ 1 }{ 2 }mV _{f}^{2}\]

OpenStudy (michele_laino):

and V is the speed of your object at initial position, namely the launch speed

OpenStudy (monsterisenergy):

Ohhhh yep that makes a lot of sense I will go over all of that once more , thanks.

OpenStudy (michele_laino):

thanks!!

OpenStudy (monsterisenergy):

Thank YOU ^-^

OpenStudy (michele_laino):

Thank YOU!!

OpenStudy (monsterisenergy):

Awh nope haha, thank YEW! xD lol.

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