Asked by lissa
A 70 kg man weighs 686 N on the Earth's surface. How far above the surface of the Earth would he have to go to "lose" 14% of his body weight?
i don't even know where to start with this question... i need help!
i don't even know where to start with this question... i need help!
Answers
Answered by
Damon
.86 = Rearth^2/r^2
because
W = G m M/R^2
and G m M does not change
remember that they really ask for
r-Rearth
the distance above ground, not the bigger radius from earth center
because
W = G m M/R^2
and G m M does not change
remember that they really ask for
r-Rearth
the distance above ground, not the bigger radius from earth center
Answered by
lissa
could try and explain more in depth? like wat does W stand for? his weight? and i don't really get how your solving it..
Answered by
Ventz
F=(G*m1*m2)/r^2
F- force
G- gravitational constant
m1- mass on earth
m2- mass of object
r- radius on earth plus given distance over it
.86*686=589.96 to find final force
plug in everything..
589.96=(6.67*10^-11*70*5.98*10^24)/ (h+6,370,000)^2
Solve for h!
h=509411
=]
F- force
G- gravitational constant
m1- mass on earth
m2- mass of object
r- radius on earth plus given distance over it
.86*686=589.96 to find final force
plug in everything..
589.96=(6.67*10^-11*70*5.98*10^24)/ (h+6,370,000)^2
Solve for h!
h=509411
=]
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.