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This answer looks at what happens to your jump height if you actually decrease in mass, while maintaining the same leg jump strength. So, let us assume that the mechanism by which a person jumps can be modelled as a finite impulse that occurs in an instance (i.e. a very very large force applied in a short frame of time). Let this impulse has a magnitude ...


0

I'm assuming that you mean you can use your thrust device further to the thrust you can impart with your own legs. Otherwise, something can lift 75 pound weight won't lift you at all. Also, you need to explain how your thrust device's force changes as you move (a spring, for example, imparts less force the more it relaxes). So I'll assume your device is ...


0

Obviously, I cant fly, but I guarantee if I weighed 75 lbs less, I could jump a lot higher than I currently can. Thank you. I appreciate your guarantee. How high could i jump if I weighed only half as much? Assuming you would be able to impart the same amount of energy with your legs, you would be able to jump twice as high. Because $$ ...


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It helps if you consider the components of the acceleration of the smaller planets due to the gravitation force of each other planet. Here is a rough diagram showing the components of acceleration for each planet, assuming the largest does not accelerate due to its large mass: The red arrow shows the component of acceleration of a planet due to the ...


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Here is the output from a program I wrote - it analyses which of the to masses between mercury (mer), venus (ven) and earth (ear) start coming closer (the third point in my question above where the heavier object is noted as making contact first - the earth is the reference ). From the below output we can see that after starting off with the same distance ...


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You raise an interesting point about the role of experiment and falsifiability in science. Despite a long-standing anomaly in Mercury's perihelion, Newton's theory of gravity itself wasn't heavily questioned, let alone rejected or falsified. Rather, auxiliary assumptions were concocted that saved Newton's theory, such as an erroneous mass of Venus, a planet ...


1

There has indeed been such research about the Pioneer anomaly: Two spacecrafts launched in the 1970s into the outer solar system did not move quite as expected (as calculated due to gravity and solar wind) after ca. 1980. Only in/after the 2000-2010 decade the source of the discrepancy, accidental thrust by thermal radiation, became generally accepted ...


1

It seems that you are mixing up the distance between the planet and the star and the distance between the star and the centre of mass (both are called $r$ in your derivation). Try to use $r = r_M + r_m$, where $r$ is the total distance between planet and star, $r_M$ is the distance between the star and the combined centre of mass,and $r_m$ is the distance ...


2

First of all, it appears that you assume the centre of mass of the two-body system lies in the middle of the path between these bodies. This, however, isn't true. The centre of mass will lie closer to the star because that is where the majority of the mass is situated. If we define $r_M$ as the distance between the star and the c.o.m., and $r_m$ for the ...


0

$$ \vec{F} = - G m \int \frac{\varrho(\vec{r}\ ') \cdot (\vec{r} - \vec{r}\ ')}{|\vec{r} - \vec{r}\ '|^3} d^3r'$$ where $\varrho$ is the mass distribution. The Integral is taken over the whole volume (or equally: the support of $\varrho$). For mass distributions, that have no 3D-volume, you have to take some Delta-terms, e.g. $$ \varrho(x,y,z) = M\cdot ...


1

Instead of integrating the force due to each mass element, which requires you to compute the component in the x-direction, you can calculate the gravitational potential, which is a scalar quantity. The force is then minus the gradient of the potential. The gravitational potential energy is: $$V(x) = -\frac{m_1 m_2 G}{\sqrt{x^2 + R^2}}$$ The force in the ...


1

Calculating the gravitational force on the axis of a ring is equivalent to calculating the gravitational force of a pair of opposing small portions of the ring in which the full mass $M$ of the ring is thought to be concentrated. The result is an axial force equal to $$F = G M m\frac{cos(\theta)}{S^2}$$ Where $\theta$ is the half-angle between the two ...


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$F=\frac{Gm_1m_2}{r^2}$ is wrong because you are essentially summing up the magnitude of the gravitational forces of each point on the ring without considering their direction. This is wrong since force is a vector. To find the force, take 2 small elements, diametrically opposite on the ring each of length $dl$ and mass $dm$. Draw the direction of the force ...


0

Two objects of unequal masses, ignoring other forces would be attracted by at the same velocity provided by the acceleration due to gravity. But objects with less mass are generally less dense, causing other agents to friction it's movement to the ground, while object which are more dense and have more mass will be more rigid to that force. This would cause ...


3

Since he is falling at terminal, hence constant, velocity, he is experiencing NO net (total) force. There are (at least) 2 forces acting on him though, which are his weight (900N) and the air resistance. Because he has constant velocity, i.e. he is not accelerating, these forces must balance, i.e. the air resistance is 900N upwards whilst his weight is 900N ...


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Finding the potential radii is actually quite simple. I already have: $$V_{eff}(r) = \frac{L^2}{2mr^2}-\frac{GMm}{r} - \frac{GMmL^2}{r^3(mc)^2} $$ I was mistakenly graphing $1/r^2 - 1/r - 1/r^3$, when in actuality it makes more sense to take $1/r^2 - 1/r - 0.1/r^3$, since $1/r^3$ is sufficiently small. Since the orbits are circular, the potential radii ...


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$dM$ is just $\rho dV$, where $\rho$ is the density and $dV$ is the volume element. In your case then $dM = \delta(r-R)\delta (\theta - \pi/2)\lambda r^2 dr d\theta d\phi = \lambda R d\phi$


1

Disregarding the constants, you have an integral in the form $$ I = \int \frac{e^t - 1}{e^t + 1}dt. $$ Now, it might be clear to apply various methods at that point. I choose to make the following: $$ I = \int \frac{e^{t/2}(e^{t/2}-e^{-t/2})}{e^{t/2}(e^{t/2}+e^{t/2})}dt. $$ By eliminating the $e^{t/2}$ factors and recognizing the terms in the ...


0

Your equation has the form $$ V_\text{eff}(r) = \frac{\alpha}{r^2} - \frac{\beta}{r} - \frac{\gamma}{r^3} $$ If you set $\alpha=\beta=\gamma=1$, then you're overestimating the $r^{-3}$ term, which is supposed to be a small correction. You will only find two extrema if the derivative has two roots: $$ V_\text{eff}'(r) = -\frac{2\alpha}{r^3} + ...


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Try to answer these. In equilibrium what balances gravitational force and centrifugal force? What is the expression for kinetic energy? What is the time period of a body in uniform circular motion with given velocity and radius of circle?


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Keep in mind the following things: Newton's Law. Newton's Law Hooke's Law.


0

Let initial velocity be $v_i$ and final velocity be $v_f$. Using 3rd equation of motion $2gS=v_f^2-v_i^2$ Since $v_f=0$ and $g=-10$ Therefore $-20*2.5=-v_i^2$ $v_i = 7 ms^{-1}$ Now using Second equation $S = v_i*t -(1/2) gt^2$ We can find t


0

Let the initial velocity when the springbok jumps be u and the velocity,v when it has achieved maximum height will be zero. Using a= -10, s=2.5 and v=0 in $$s=vt-\frac { 1 }{ 2 } g{ t }^{ 2 }$$we can find time t. Using a= -10, t=0.707 and v=0 in $$v=u+at$$we can find the initial velocity u.


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It is possible to find $t(r)$ and $t(\theta)$ easily, however inverting to find $r(t)$ and $\theta(t)$ is hard to do in general (unless you use Fourier series etc..) \begin{align*} t(r) &= \sqrt{\frac{m}{2}} \int \frac{dr}{\sqrt{\frac{k}{r} - \frac{l^2}{2mr^2} + E}} \\ t(\theta) &= \frac{l^3}{mk^2} \int \frac{d\theta}{\left( 1 + e \text{cos }(\theta ...


1

But I am wondering if it is possible to solve the two equation of r with respect to t and θ with respect to t. Yes, you can. In the case of an elliptical orbit with non-zero angular momentum, you need to introduce the concepts of eccentric anomaly, mean anomaly, and mean motion. Eccentric anomaly is related to true anomaly (your theta) via ...


1

No. Most of the kinetic energy from a rocket is sideways (8 km/s) so as to let the spacecraft enter orbit. So if you want your orbital airplane, you need an engine that can get up to mach 25 inside of the atmosphere (so you can use relatively efficient airbreathing engines), you need: 1: a high-speed airbreathing engine (a scramjet) 2: a thermal protection ...


5

Your question highlights a common misconception. A satellite in orbit around the Earth is accelerating towards the Earth right now. Any object moving in a circular path has an acceleration towards the center of the circle because the direction, and therefore the velocity, of the object is constantly changing. This acceleration, called centripetal ...


0

If they both were in orbit, then they would already be accelerating down towards the earth, so there wouldn't be anything sudden about it - they have the exact same acceleration before and after impact.


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Absolutely we could, and in fact, I strongly suspect that General Relativity was never used in the Apollo program. for one thing, the on-board navigation computers were nowhere near powerful enough to perform any useful calculation with GR. on the other hand, it's possible to measure the position of the moon to within a few centimeters (much more accurate ...


3

Consider that it would not be particularly difficult to do an Apollo-type landing if each of your relative velocity, range, and angular measurements were off by +/- 5%. You could simply make small iterative corrections along the way, until the absolute values were small enough to make the relative errors inconsequential. At worst you'd need to carry ...


1

Writing it the way you did, you don't try to compensate the force, but to set it equal. This means you account for the fact that the centripetal force is gravity and nothing more and they do point in the same direction for a circular orbit. Assume a body in circular motion around the origin of a coordinate system at distance $R$. One can express this motion ...


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The Jet Propulsion Laboratory has incorporated general relativistic effects in its numerical integration of the planets since the mid to late 1960s. For example, the JPL DE19 ephemeris, released in 1967, incorporated relativistic effects in its modeling of the solar system. This didn't help much. Had they ignored relativistic effects there would have been ...


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A few sanity checks without actually computing anything: First, the error due to neglecting general relativity is so small that it didn't affect prediction of lunar eclipses and wasn't actually noticed anywhere except in Mercury's orbit (at least not until they purpose-built experiments to detect minor discrepancies). I know this doesn't give a completely ...


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The trouble with orbital mechanics is that it rapidly gets exceedingly complicated and hard to make intuitive sense of. However I think there is a reasonably straightforward way to show how little effect GR has on an Earth-Moon transfer orbit. But this takes a little preparation so bear with me while I give a short introduction. I hope everyone who reads ...


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I'll start the ball rolling on this one. My GR knowledge is probably not good enough to make this a truly satisfying answer... The gravitational acceleration for an object moving radially at non-relativistic velocities in the Schwarzschild metric is modified by a factor $(1 - r_s/r)(3[1-r_s/r] -2)$, where $r_s = 2GM/c^2 = 0.00885 m$ for the Earth. If we ...


2

You are writing (apart from a minus sign) the tidal tensor, which is the traceless part of the Hessian of gravitational potential, for a point mass in the origin. This tensor contains information about the piece of gravitational force which can't be removed by choosing a free falling reference frame. I'm not sure to understand your problem, however the tidal ...


3

Suppose you start with a proton and an electron separated by a large distance. The mass of this system is just $m_p + m_e$. Now let the proton and electron fall towards each other under their mututal electrostatic attraction. As they fall they will speed up, so by the time the proton and electron are about one hydrogen atom radius apart they're moving with ...


1

Because E = mc^2, B/c^2 = m. The binding energy ends up adding mass to the system of the atom.


1

For systems involving central forces, the orbit can be any of the conic sections. And which conic it will be is determined by the total energy of objects revolving around the centre of force. If the total energy is negative ( as in the case of planets), the eccentricity of the orbit will be either zero or one. If eccentricity is equal to zero, it will be a ...


2

The mistake in your reasoning is that the mass elements that are closer to you (on the closer side from the center) exert more gravitational force than those that are far away. So it is unjustified to treat the mass distribution as a mass sitting at the center. Imagine for instance a linear mass of infinite length, then according to your intuition, the ...


3

The spheres are supposed with homogeneous mass distribution, and also the force of gravity is supposed homogeneous, therefore the CoM coincides with the center of gravity and the center of the sphere, and you can just consider the distance between them: 1 kg --20cm--2 Kg-----80 cm ----- 5Kg The formula to find the center of mass $\vec{c}$ is: $$ \vec{c} = ...


0

The value of g (in your case 8ms^-2) of a planet depends on his mass m, radius r (half diameter) and the gravitational constant G. g=G*m/r^2 your 2 cases: 1) g_1=8 ms^-2 --> 8=G*4M/(2d/2)^2 --> GM/d^2=2 (take this for the second case too) 2) g_2=GM/(d/2)^2 --> g_2=4GM/d^2 --> g_2= 4(GM/d^2)=4*2=8 ms^-2 both are equal --> correct answer C


0

Although this is not a homework-problem website, let me answer anyway. $$F=ma=\frac{GmM}{r^2} \implies a = G\frac{M}{r^2} $$ $$ G\frac{4M}{d^2} = 8 \text{ m s}^{-2} \implies G \frac{M}{d^2} = 2 \text{ m s}^{-2} \implies G \frac{M}{(d/2)^2} = G\frac{4M}{d^2} = 8 \text{ m s}^{-2}$$ I think the answer should be C if I didn't make any silly algebra mistake, ...


0

Keep in mind That g ~ M and g ~ 1/R^2. So in your problem the mass of planet Y is 4 times smaller than the mass of planet X. Following this the g of Y should be 4 times smaller : 8/4 = 2. But it's radius is also 2 times smaller so because g~1/R^2, you should mulltiply it by 4 : 2*4=8. Finally The gravitational acc. of planet Y is 8m/s^2 (the same as planet ...


1

Maybe I am missing something simple, but mass divided by radius squared is the same in both cases, so I would vote for (C).


3

Dark matter does not readily "accumulate". If(?) it exists then it interacts very weakly with normal matter and is primarily influenced by gravity. The Earth's gravity is far too small to make a local concentration of dark matter. The local dark matter would be moving in the Galactic potential at speeds similar to that of the Sun around the Galaxy ($\sim ...


3

There isn't a simple closed form expression for the distance as a function of time in general relativity. However if you're just interested in how big the difference is I think there is a nice way to see this. For simplicity let's take a falling object with zero total energy. In the Newtonian case this means the kinetic energy is equal to the potential ...


-2

I believe I've heard Neil deGrasse Tyson say its roughly 17 miles per second to escape Earth's gravitational force in a spacecraft like the Apollo rockets that were used. So I'm assuming we're basically talking about how fast something has to be accelerating to escape Earth's mass/gravity. I believe that is correct - 17 miles per second.


2

It would be more convenient to use the fact that $F\Delta t = m\Delta v$. There is a constant force of gravity pulling you down, and an occasional force of the foot pushing you back up. Averaged over time, the two must be equal as there is no net change in vertical velocity. This means that $F_{strike} \approx \frac{W}{\Delta T}$ where $W$ is the weight and ...


0

The difference is often due to the addition of the Lorentz factor, which you can apply to equations of motion using a Lorentz transformation. Dr. Wolfgang Rindler was kind enough to write a small paper on this on scholarpedia. The Lorentz factor ($\gamma$) is often defined as: $$\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$$ This changes basic momentum from ...



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