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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 ...


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 ...


31

Both values are computed as a position weighted average. For the center of mass we average the mass in this way, while for the center of gravity we average the effect of gravity on the body (i.e. the weight). $$ \begin{align*} x_{com} &= \frac{\int x \, \rho(x) \,\mathrm{d}x}{\int \rho(x) \, \mathrm{d}x} \\ \\ x_{cog} &= \frac{\int x \, \rho(x)\, ...


23

As a quick rehash of layman's terms definitions you've probably heard: The Center of Mass (CM) represents a single point where you could treat the object as a point particle, with the combined mass of the object. It's found by the average location of the mass of an object. The Center of Gravity (CG) is a point that represents the average pull of gravity on ...


23

The main plot below shows the potential energy of a mass in the Earth-Moon system under the unrealistic assumption that the system is not rotating. i.e. This mirrors (at present) all but one of the 4 answers given, in assuming that this point is defined where the gravitational force on a mass due to the Earth and the Moon are equal and opposite (i.e. at the ...


16

Set the forces on the test particle from the Earth and Moon equal: $$F_E=F_M$$ $$G\frac{M_EM_{\text{ test particle}}}{R_E^2}=G\frac{M_MM_{\text{ test particle}}}{R_M^2}$$ The $G$s and $M_{\text{ test particle}}$s cancel, leaving you with $$\frac{M_E}{R_E^2}=\frac{M_M}{R_M^2}$$ but you know that $R_M$, the distance between the test particle and the Moon, is ...


12

Just based on the quadratic drag of air, yes, the fired bullet would take longer to hit the ground. Just consider the vertical force caused by the air friction: $F_y = - F_{\rm drag} \sin \theta = - C (v_x^2 + v_y^2) \frac{v_y}{\sqrt{v_x^2 + v_y^2}} = - C v_y \sqrt{v_x^2 + v_y^2}$ Where $\theta$ is the angle above the horizon for the bullet's velocity, ...


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 ...


10

Yes, Newton's formula is just fine. No, the formula in your book doesn't describe reality. At first this sounded like an exercise, where the next sentence is probably something like "calculate the effect this has..." These sorts of hypothetical questions are meant to show you how you could distinguish between competing physical theories. Some more digging ...


10

One intuitive way I've seen to think about the math is that if you are at any position inside the hollow spherical shell, you can imagine two cones whose tips are at your position, and which both lie along the same axis, widening in opposite direction. Imagine, too, that they both subtend the same solid angle, but the solid angle is chosen to be ...


9

Earth is about 100x more massive than the moon, and since $F \propto M / r^2 $, the distance from Earth to the astronaut would have to be about $\sqrt{100}$ = 10x further than from the moon to the astronaut. Therefore, the astronaut falls "up" about 90% of the way to the moon. [The earlier answers go a lot more into detail (and are more technically ...


9

At Lagrange point L1. Specifically for Earth-Moon L1, these calculations show 326054 km.


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.


8

The centre of mass is the average point of the "mass" of the body, whereas centre of gravity is the average point of the "weight" that is mass times the local gravitational acceleration. For small objects both of them are almost same, but for large objects as the value of gravitational acceleration can change along the body (as the gravitational ...


5

According to the research paper linked here: http://www.researchgate.net/profile/Weicheng_Cui/publication/222221948_An_overview_of_buckling_and_ultimate_strength_of_spherical_pressure_hull_under_external_pressure/links/53f1a2950cf26b9b7dd0da3c The pressure difference which can be held by a sphere of any particular material is a function of (t/R), where t is ...


4

For your 2 minute egg timer here on Earth it comes out to be 4 minutes 54 seconds on the Moon because: $t_{Moon} = t_{Earth} \sqrt{6}$ Full explanation below. Q: What is the relationship between hourglass flowrate and local gravity? As in the excellent answer to a related question (hourglass flowrate vs. sand grain size) and this published paper, the ...


4

You are correct that as you get very close to the center of the earth, the value of $g$ can become arbitrarily low. If you could somehow create a space there, you could potentially float in it because you would not be pulled in any particular direction with respect to the earth. But while gravity is not strong there, it is strong in other places (like your ...


3

The distance I got was 346 084km. Here are the maths I used: ($E_m$) Earth mass = $5.9736\times10^{24}$ kg ($M_m$) Moon mass = $7.3477\times10^{22}$ kg ($D_{em}$) average Earth-Moon distance = 384 467km ($G$) gravitational constant = $6.67384\times10^{-11}$ ($W$) my weight = 85kg ($D_{fe}$) distance from earth = ? The attraction force between two objects ...


3

To calculate this by yourself, you need to know that gravity force exerted on an object (for exapmle You) is equal to $F=GMm/r^2$, where $G$ is gravity constant, $M$ is the mass of the big object ($M_m$ for moon, $M_e$ for earth), $m$ is the mass of small object. $r$ is the distance from the center of the mass. Now you need to know masses of earth and moon ...


3

Just use an equation derogating from the two forces which pull the objects (universal gravitation)to get the equilibrium point, something like (already simplifyed): M/d^2 = m/(384000000 - d)^2 Where M is the mass of earth, m the mass of moon and d the distance from earth. As d gets bigger than this value, you start falling into the moon I get a value of ...


3

Your equation relates the period of the orbit to the length of the semi-major axis, not to the absolute distance at any point. You can use the Vis-viva equation if you have more information. But you don't have the semi-major axis length or other details about the orbit. As you suggest, conservation of energy is the simpler way forward.


3

For a heavy enough hammer, and not too high distances, the friction from the air will not be too large, because it is a function of speed, so the smaller the speed and the higre the mass the smaller the effect of the air friction. But even if there is some friction this should not be a problem (as I will explain at the end). If in your experiment the cage ...


2

1) What is the initial position of the bannana? 2) invent an initial position for the shooter. 3) at time 0, what is the direction of the bullet's velocity? 4) Now, set up a system of equations for the bullet and the bannana's motion. 5) do they hit each other?


2

It seems to me you may be misunderstanding the problem as stated. You are assuming you are being asked about two different objects (planets?) in different orbits; but I think from reading the question that you are being asked about the same object at different points in its elliptical orbit. For an object in an elliptical orbit, conservation of angular ...


2

In principle not, because it is wrong when describing the behavior of the orbit of Mercury, but should be borne in mind that there is no absolute truth when describe the universe, they always talk about "good approximations" and Newton's law of gravitation rule! xD, (it can take you to the moon!).


2

Clearly missed the point in this statement. For both the 1000kg and 1kg masses, the product of them both with the earth's mass is clearly the earth's mass, so F will be virtually the same for them both. The product is definitely not the same and the force on 1000kg ball is exactly 1000 times greater that the force on 1kg ball and much much more on the ...


2

The gravitational force between two bodies acts in a straight line between the two. So in this case it is $F = G \frac{m \cdot M}{r^2+R^2}$ (using Pythagorean theorem for the distance)


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 ...


2

It is true that the shell outside the current radius does not contribute, so you are only left with the force from the mass inside: $$F=G\frac{mM_{inside}}{r^2}$$ but the M inside becomes smaller as the radius becomes smaller.Assuming the dendity, $\rho$, is a constant, then $$M_{inside}=\rho\frac{4}{3}\pi r^3 $$ which leaves you with: ...



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