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5

Consider three bodies $Earth$, $B$ and $C$. Now, we take two cases. In the first case, we will observe gravitational attraction between body Earth and the heavier body ($B$); and in the second case, we will observe the gravitational attraction between body Earth and the lighter body ($C$). Case 1 Consider body $Earth$ and $B$, separated by a distance of ...


0

If you assume that the earth is immobile, the fall time if the same. If you consider that the earth moves because of the mass of the item, yes, the heavy item fall time is shorter (infinitesimal difference). If both items are dropped at the same time, they will hit the floor at the same time though. (This question has already be answered : Don't heavier ...


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 ...


1

The key to understanding this is the phrase The potential part derives from the combined height of the center of mass of the falling dominoes, for which we take the dimensionless quantity Followed by the equation: $$H_n(\theta_n)=\sum_i^n{\left[\cos\theta_i + (d/h)\sin\theta_i\right]}$$ In other words - this is describing the equivalent height of ...


0

yes it would, (well if it was moving down a ramp) like you said, friction would also come to play, but it should go faster because the more weight you add the more downwards momentum and if it's moving dow a ramp then it should turn that to forward momentum.


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 ...


0

I'm probably wrong, but I think it would take about 6 times as long, assuming the moon's surface gravity is 1/6th that of Earth: According to http://www.physics.umd.edu/deptinfo/facilities/lecdem/services/demos/demosc5/c5-41.htm, "[d]uring the steady-state sand fall the extra force of sand hitting the bottom very nearly cancels the loss of weight of ...


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 ...


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 ...


22

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 ...


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 ...


9

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


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 ...


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 ...


0

Applying simply the classical mechanics, the centrifugal force rω^2 should be equal to the gravitational acceleration GM/r^2, where r is the radius you look for, ω is the angular velocity, and M is the mass of the earth. So, rω^2 = GM/r^2 ==> r^3 = GM/ω^2 . The angular velocity ω of the earth rotation can indeed be calculated from the fact that in ...


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 ...


1

An interesting question, but I think this would be absolutely impossible for the following reason. Vacuum chambers, particularly large ones, need thick strong walls to prevent air pressure making them collapse. The larger the volume the greater the problem. Pressure = force over area - so force = area times pressure - $F=PA$ For area of $10 m^2$, Force ...


0

Wether the car will land on its 4 wheels or not is subjective to the cars centre of mass and the height of the cliff. As soon as 2 of the car's wheel are of the cliff , the car is not in equilibrium. Therefore the car will tilt toward the front and then depending on its centre of mass tip over and complete an entire flip. The time taken for an entire flip of ...


0

In general a two-axle vehicle driving off of a abrupt edge will tumble back-over-front on the way down. Which way it lands therefore has to do with how fast it tumbles and how long it tumbles for. The reason for the tumbling is because for the time it takes to drive the distance between the axles, the back axle experiences the normal upwards force from the ...


1

No, cars/vans generally will land front-first or on the roof. This is due to the fact that once the front-end of the car starts going over the edge, a torque (due to gravity) induces a rotation on the vehicle. Most highways and bridges around are small enough that the car/van can only rotate a quarter or half turn (which corresponds to the front and roof ...


0

Not in general. Think of it like this: when the car drives off the cliff, at some point it has a portion of it falling and consequently rotating due to torque exerted by gravity while the other end remains at its original height due to normal forces. As a result, the car will tend to have some sort of rotation as it falls. Thus, the height of the cliff ...


1

One way to think of this conceptually is that the inertial mass of the system is in fact different from the gravitational mass. That is, while the net force on the systems are the same, the two systems have a different amount of mass, and so they resist changes in velocity to different degrees. Except that in this case, we are considering a compound ...


2

Assuming that you want to replicate David Scott's Apollo 15 version of Galileo's Tower of Pisa thought experiment, I think that something like your experiment could indeed work. As Martin says, the boxes would need to be very low drag shapes: presumably you could make them out of a hollow very heavy metal - natural or depleted uranium would probably make ...


2

Gravitational potential plus the centrifugal potential due to the Earth's rotation at the Earth's surface is approximately given by $$\begin{aligned} U(r,\theta) = -\frac {\mu_e}{r} + \frac {\mu_eJ_2 a^2}{r^3}\frac{3\sin^2\theta-1}{2} - \frac 1 2 \omega^2r^2\cos^2\theta &\qquad&\qquad(1) \end{aligned}$$ where $\mu_e = GM_e$ is the Earth's ...


2

Well, a circle is a special case of an ellipse, so if something is true of a general ellipse, it will be true of a circle. In order to work out the rules for ellipses, the calculus is typically a bit more involved than what you see in your standard intro class, so the real, general case is usually reserved for sophomore year classical mechanics, where you ...


7

I was revisiting triple integrals, so I decided to give this a go. This is the result: As expected, the pole wins. Development Starting from Newton's law of universal gravitation: $$\mathrm d\mathbf g = G\frac{\mathbf r\ \mathrm dm}{\left|\mathbf r\right|^3}$$ And the parametrization of our oblate asteroid: $$(x,y,z) = \mathbf ...


1

In the flat, you will feel it falls down the mountain side to the center Due the direction of gravity would be at the center of mass. A picture is worth a thousand words. In a uniformly mass distributed hemisphere of radius r, the center of mass is given by geometric centroid. The geometric centroid of an heisphere is then given by 3/8 * r ...


2

The appeared weight of an object does not only depend on the mass of the celestial body by which it is attracted. If you simplify to spherical symmetry, which is not definitely not that case for the comet 67P (in to a lesser extend also not for the Earth) you can approximate the ratios of weight by using Newton's law of universal gravitation: $$ ...


7

Using the same notation, assumptions and approximations as my Phys.SE answer here, the gravitational potential (monopole + quadrupole) is $$U(r) ~\approx~ U_1(r)+U_4(r) ~\approx~ -\frac{GM}{r}\left(1 + \frac{(3s^2-2)fR^2}{5r^2}\right).$$ Here $a$ and $b$ are the equatorial and polar radius of the oblate spheroid; $$0~<~f:=~1-\frac{b}{a} ~\ll~ 1$$ is a ...


-1

For when the guy stands on the pole, slice the planet into circular disks, with their centers on the axis, each with radius = r = f(h) and thickness = dh, where h is distance of the center of the disk from one pole, and f(h) is function that represents shape of oblate spheroid. Then slice each disk into multiple rings, nested one within the other, each ring ...


2

As you rightly pointed out, the fact that 67P is oddly-shaped should alter its gravitational attraction on various parts of the comet. That said, if we were to go by Wikipedia in a rather off-hand manner, we find that the lander is $100 kg$ (as @fibonatic rightly pointed out) and 67P has an acceleration due to gravity of $\textbf{g'} = 10^{-3} m/s^2$. Its ...


0

If the density of that oblate spheroid is constant then gravitational force is bigger in Case 1 because in that case the person stands closer to the center of the mass.


1

There is a recent paper in which $G$ has been measured. We obtain the value $G = 6.67191(99) × 10^{−11} m^3 kg^{−1} s^{−2}$ with a relative uncertainty of 150 parts per million (the combined standard uncertainty is given in parentheses). Most measurements of $G$ are based on a torsion pendulum as in the original Cavendish experiment. The paper ...


0

Consider that, The force of gravity falls off with distance (the inverse square law). Gravity is a property of the components, not the whole. So it is the atoms of your oblate spheroid that are exerting gravity on the person (and these forces sum over the whole object) and one way to work this out is to calculate the force of gravity exerted by each atom ...


1

This reminds me of the Apollo 15 mission where a similar experiment was preformed on the moon. Outside a vacuum, you would have to encase the feather in a reasonably aerodynamic container in order to compensate for air resistance.


3

Using a big vacuum chamber air resistance is removed. There is a video here.


0

That completely depends on the distance Earth-Sun. If the distance is high enough (currently no numbers, but I can add some later), then we can feel something, but the gravity difference is so low that only instruments can measure the difference. The nearer we get, the more we are going to feel the difference. First the people directing towards sun are ...


2

Like mentioned in the comments, you got the initial velocity wrong. You source says: Further to this, the selected separation strategy foresees a fixed separation velocity of approximately 0.187 m/s. This imposes limitations on the capability of directing Philae towards the comet, i.e. it restricts the domain of possible positions and velocities ...


-2

Well, we have people expounding on the equilibrium being unstable but, uh, wouldn't the net force result in an attraction between the centres of mass? To me, that seems like an essentially stable setup, at least before taking rotation into account.


5

If the ring elements are actually in orbit (i.e. each part would follow the ring shape by itself without need for a pillar), then you get this: If each ring element is "closer" to the Earth (i.e. they don't revolve fast enough for their altitude), then the pillars offer sustenance. If you remove them, then the ring is stable in the north-south axis, but ...


1

The responses so far seem to suggest that, assuming known density, we can tell how far away an object is by the strength of its gravitational attraction. I'm quite sure this is wrong. Since we are in free-fall about the sun (as we are about the moon) I don't think there is any direct way to measure the force of its attraction. It is however possible to ...


0

For the two orbs, we use Newtons universal law of Gravitation: $$ F_{1\to2}=G\frac{m_1m_2}{r_{12}^2}\tag{1} $$ which does not require any information about densities, purely the total mass of the objects. The data we need are: $m_1=5.972\times10^{24}$ kg $m_2=7.34767309 \times 10^{22}$ kg (moon) $m_2=1.99\times10^{30}$ kg (sun) $r_{12}=384,400$ km ...


5

The apparent diameter of an object like the Sun or Moon varies as $r^{-1}$, that is an object twice as far away looks half the size. So the nearly perfect fit during an eclipse tells us that the Earth-Moon distance $d_M$ and the Earth-Sun distance, $d_S$, are related to the radius of the Moon, $r_M$, and the radius of the Sun, $r_S$, by: $$ \frac{d_M}{d_S} ...


3

The motion of a satellite in orbit is governed by Conservation of angular momentum Conservation of energy When there is no (or negligible) drag, these two are satisfied by an elliptical orbit. Kepler studied this at length, by observing the motion of the planets. He formulated his famous laws: The orbit (of a planet) is an ellipse (with the sun at one ...


4

If you solve the equations for two bodies interacting via gravity (or indeed any inverse square law force) then the bound orbits are all elliptical - a circle is a special case of an ellipse with zero eccentricity. So any object in an elliptical orbit will remain in that orbit forever. No external forces are needed. To see how the orbit is calculated have a ...



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