Asked by albert
find the gravitational force exerted by the sun on a 70 kg man located at the earth's equator
a)at noon when the man is closest to the sun.
b)at midnight when he is futherst from the sun.
c)calculate the diferrence in gravitational acceleration due to the sun between noon (a) and midnight (b).
d)explain how this differencein (c) above causes oceantides for the case of the moon.
a)at noon when the man is closest to the sun.
b)at midnight when he is futherst from the sun.
c)calculate the diferrence in gravitational acceleration due to the sun between noon (a) and midnight (b).
d)explain how this differencein (c) above causes oceantides for the case of the moon.
Answers
Answered by
Damon
I assume you can do
F = G Msun Mman / R^2
but the tide is more complicated and depends on the moon. Look at:
http://scijinks.jpl.nasa.gov/tides/
F = G Msun Mman / R^2
but the tide is more complicated and depends on the moon. Look at:
http://scijinks.jpl.nasa.gov/tides/
Answered by
Damon
by the way the difference in R between one side of earth is tiny compared to R
If you have had calculus think:
F = k/r^2
dF/dr = -2r /r^2
so
dF = (-2/r)dr
so you can calculate a small change in F due to a small change in r
If you have had calculus think:
F = k/r^2
dF/dr = -2r /r^2
so
dF = (-2/r)dr
so you can calculate a small change in F due to a small change in r
Answered by
Damon
dF/dr = -2rk /r^2
so
dF = (-2k/r)dr
so
dF = (-2k/r)dr
Answered by
Damon
dF/dr = -2kr/r^4
= -2 k/r^3
so
dF = (-2 k /r^3) dr
= -2 k/r^3
so
dF = (-2 k /r^3) dr
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