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4

As explained in this article by Neill DeGrasse Tyson, the tidal forces between the Earth and the moon do indeed slow down the rotation of the Earth each year, the same process that caused the moon's rotation to become tidally locked with its orbit of the Earth. This effect would eventually cause the Earth's rotation to be tidally locked with the moon as ...


11

As stated in other answers it is how much the gravitational force is different by on opposite sides of the earth that creates the tides. You can still show this using $a=\frac{GM}{d^2}$ but you need to consider the difference, not the absolute force on the earth. The sun while much more massive is just far enough away that it is getting to a much flatter ...


8

The highly upvoted answer is right but to make things much simpler: Tides are based on the change in gravity, not the gravity. That means they drop off at the cube of the distance rather than the square of the distance like gravity itself does. Thus the object with the most gravity isn't necessarily the one that causes the most tide.


35

Tides are caused by the gradient of the gravitational field - so the tidal "force" experienced drops with the third power of the distance. This means that the relative strength of the tides should go as $$ratio = \frac{M_{moon} \cdot D_{sun}^3}{M_{sun} \cdot D_{moon}^3}\\ =\frac{7 \cdot 10^{22}\cdot (1.5 \cdot 10^{11})^3}{2\cdot 10^{30}\cdot (3.7\cdot ...


61

What is important for tidal forces is not the absolute gravity, but the differential gravity across the planet, that is, how different the gravity is at a point on the surface near the sun relative to a point on the other side. Because the Sun is far away, gravity doesn't change much between the two extremes on earth. However, if you compare it with the ...


0

As shown in a previous answer, Assuming they have the same density (the Sun's average density is not much smaller than that of the moon) , if they had the same apparent size in the sky, then the mass M of the object will grow as r3 (because M=4/3ρπR3 and R=θr), so the force actually grows linearly with r. this implies that for same apparent size and ...


0

The gravitational tides generate atmospheric tides ( http://en.wikipedia.org/wiki/Atmospheric_tide ), earth tides and water tides. The water tides can be found in oceans, groundwater or rivers (about the last 2 types you can read on http://www.nature.com/srep/2014/140226/srep04193/full/srep04193.html ). The moon is the principal cause of the tides on Earth; ...


2

Tidal forces drop rapidly with distance - the derivative of $1/t^2$ is $-2/r^3$. Further, the difference in radius of the orbits of Earth and Mercury is a little more than a factor 3x and radius of mercury is about 2.5x smaller than that of earth. From the orbits we gather the tidal effect is 27x smaller - from the radius we gather that moment of inertia is ...


1

time slows in the presence of strong gravitational fields It's not the gravitational field that determines time dilation, it's the gravitational potential. The Newtonian approximation really isn't correct here, but let's use it anyway for insight: The potential falls off like $1/r$ with distance $r$. The field falls off like $1/r^2$. Tidal effects go like ...


2

Rosetta initially went round the comet in a roughly triangular orbit using its thrusters to change direction. Eventually in September 2014 it entered a true orbit at a distance of about 30km going round once every two weeks. The orbit can hold in such weak gravity because it is very close and slow compared to satellites orbiting Earth. It has now moved ...


1

A stable orbit requires that the orbital insertion velocity be just right. Too high and the spacecraft flies away on a hyperbolic trajectory. Too low and it eventually falls and impacts the planet (or comet in this case). Upon approaching a planet or comet the spacecraft will perform a delta-v maneuver, a burning of thrusters that insert the spacecraft into ...


5

First off, we should take the wisdom of Jack Wisdom to heart: "Calculation of the history of the lunar orbit is fraught with difficulties." While calculating the history may well be fraught with difficulties, calculating the future is hugely problematic. For the sake of argument, I'll ignore that issue. The OP suggests the end of the planet is about 5 ...



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