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.
dF/dr = -2rk /r^2
so
dF = (-2k/r)dr
dF/dr = -2kr/r^4
= -2 k/r^3
so
dF = (-2 k /r^3) dr
a) Well, if the man is closest to the sun at noon, he must have some serious sunburn! Anyway, to calculate the gravitational force exerted by the sun, we can use Newton's law of universal gravitation. The formula is F = (G * m1 * m2) / r^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the objects (in this case, the man and the sun), and r is the distance between them. So, let's plug in the values! Just hold on my over-sized clown calculator... *beep boop beep* Okay, so the gravitational force exerted by the sun on the man at noon would be... um... approximately 0, because the gravitational force between the man and the sun is negligible compared to the force of gravity between the man and the Earth. Don't worry, the man won't float away!
b) Now, if the man is farthest from the sun at midnight, he can finally relax and enjoy the night sky! To calculate the gravitational force in this case, we still use the same formula. Calculating... *beep boop beep* Ah, it seems that the gravitational force exerted by the sun on the man at midnight would still be approximately 0. Again, the gravitational force between the man and the sun is pretty tiny compared to the force of gravity between the man and the Earth.
c) Since the gravitational force exerted by the sun on the man is practically the same at noon and midnight, the difference in gravitational acceleration due to the sun between these two times is also negligible. Well, this means that the gravitational force from the sun has little impact on the acceleration of the man.
d) As for the ocean tides caused by the moon, it's a totally different story! The moon's gravitational force is much stronger than the sun's force on Earth because it is closer. The tidal force is created when the moon's gravity pulls more on the water that is closer to it compared to the water on the opposite side of the Earth. This difference in gravitational force causes a tidal bulge, resulting in high and low tides. So, the moon is the real star when it comes to ocean tides, not the sun. The sun just enjoys a sunny day at the beach!
To find the gravitational force exerted by the sun on a 70 kg man located at the Earth's equator, we can use Newton's law of universal gravitation:
F = G * (m1 * m2) / r^2
where:
F is the gravitational force
G is the gravitational constant (approximately 6.67430 × 10^-11 m^3 kg^-1 s^-2)
m1 is the mass of the man (70 kg)
m2 is the mass of the sun (approximately 1.989 × 10^30 kg)
r is the distance between the man and the sun
a) At noon when the man is closest to the sun:
The distance between the Earth and the sun is not constant due to the Earth's elliptical orbit. However, for simplification purposes, let's assume the average distance from the Earth to the sun (also called astronomical unit or AU) is about 149.6 million kilometers (approximately 1.496 × 10^11 meters).
Using this average distance (r) in the formula, we get:
* (m1 * m2) / r^2
= (6.67430 × 10^-11) * (70 * 1.989 × 10^30) / (1.496 × 10^11)^2
b) At midnight when the man is furthest from the sun:
Again, let's assume the average distance to the sun at its farthest point is approximately 152.1 million kilometers (approximately 1.521 × 10^11 meters).
Using this average distance (r) in the formula, we get:
F(b) = G * (m1 * m2) / r^2
= (6.67430 × 10^-11) * (70 * 1.989 × 10^30) / (1.521 × 10^11)^2
c) To calculate the difference in gravitational acceleration due to the sun between noon (a) and midnight (b):
Gravitational force and gravitational acceleration are related through Newton's second law:
F = m * a
We know that F = G * (m1 * m2) / r^2 from above, and we can rearrange this equation to solve for gravitational acceleration (a):
a = F / m
Using this equation, we can calculate the gravitational acceleration for both noon and midnight, and then find their difference.
d) Regarding ocean tides caused by the Moon:
Ocean tides are primarily caused by the gravitational pull of the Moon rather than the Sun, because the Moon is much closer to the Earth. The Moon's gravitational force causes the ocean water to bulge towards it, creating two high tides on opposite sides of the Earth.
The difference in gravitational acceleration between noon and midnight due to the Sun (c) does not play a significant role in ocean tides. However, the combined gravitational forces of the Sun and Moon at different positions relative to the Earth do affect the tides. When the Sun and Moon are aligned, their gravitational forces combine, leading to higher high tides (spring tides). When the Sun and Moon are at right angles, their gravitational forces partially cancel each other, resulting in lower high tides (neap tides).
In summary, the gravitational force exerted by the Sun on a person at the Earth's equator can be calculated using Newton's law of universal gravitation. The difference in gravitational acceleration between noon and midnight due to the Sun does not directly cause ocean tides, but the combined gravitational forces of the Sun and Moon do influence the tides.
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/
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