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Physics 17 Online
OpenStudy (anonymous):

A spacecraft traveling from Earth to the Moon reaches a point where the pull of the Moon’s gravity is equal to that of the Earth’s. Before this point the speed of the spacecraft is decreasing and after this point the speed is increasing (a little like coasting up a hill and then back down the other side). Determine the precise location of this point.

OpenStudy (michele_laino):

the condition that characterizes the requested point, is that the gravitational force exerted by the earth on the spacecraft and the gravitational forcr exerted by the moon on the spacecraft have the same magnitude. So we have to solve this equation: \[\Large K\frac{{{M_T}{m_s}}}{{{x^2}}} = K\frac{{{M_L}{m_s}}}{{{{\left( {d - x} \right)}^2}}}\]

OpenStudy (anonymous):

what does K and x stand for??

OpenStudy (michele_laino):

where K is the Newton constant, M_T and M_L are the masses of the earth and moon respectively, m_s is the mass of our spacecraft, d is the mean distance earth-moon, and x is as below: |dw:1429675228093:dw|

OpenStudy (anonymous):

so k=6.67 x 10^-11?

OpenStudy (michele_laino):

that's right!

OpenStudy (anonymous):

this looks way more advanced than what we've been doing but I'll try to follow along :)

OpenStudy (michele_laino):

ok! :)

OpenStudy (anonymous):

how can I carry out this equation if I dont have the mass of the spacecraft?

OpenStudy (michele_laino):

we don't need to know the mass of the spacecraft. now, after a simplification, we can write: \[\Large {\left( {\frac{{d - x}}{x}} \right)^2} = \frac{{{M_L}}}{{{M_T}}}\]

OpenStudy (michele_laino):

|dw:1429675625673:dw|

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