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We'll start with your second question, as that's (much) simpler. Close to the surface of the earth, it's safe to assume that the force of gravity is proportional to the mass of your object: $$F_G = mg$$ where $m$ is the mass, $F_G$ is the force of gravity, and $g$ is a constant (for the earth $g \approx 9.8 m/s^2$.) Then Newton's second law tells us that ...


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If there is no resistance then there will be no net torque applied about the center of mass. So any initial rotational speed will remain. The rotation center is going to be the center of mass. The effect of gravity will be to accelerate the center of mass, and it will have no effect on the rotational motion of the body. See this accepted answer for a ...


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Ignoring the Coriolis effect, which is velocity dependent. I get for an x-y Cartesian coordinate system, with the y component toward the North Pole and x component pointing out from the equator at the position you are measuring the effective g=G , the following (in ordered pair vector representation); G= ...


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There are private space companies that will allow you to take a trip to space(it costs around $250,000). Take a trip and go in orbit around the earth or go to another space station. Another solution would be to go to the Zero-G place as Lasse said or fall farther as Rob Jeffries said which are both much less expensive.


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Establish a longer free fall time, for example in an aircraft or space station.


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Who is right? Galileo. This sounds a little like a homework problem. Is it? If so it should be labelled as such. The sun rises in the east and sets in the west. Thus the earth is rotating "to the east". The earth ostensibly has a constant rotational velocity $\omega$ of about one revolution per 24 hours. That means a point at the base of the tower has ...


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When the ball is being held "stationary" at the top of the tower, then it is moving with the same angular velocity as the Earth. However, this does not mean that its angular velocity will remain constant during the drop. Indeed, it can't: because the object is moving at a nonzero radius from the axis of the Earth's rotation, it has angular momentum, and that ...


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If only the forces of gravity are present, all objects fall at the same rate. This is what one calls equivalence principle. In classical mechanics it shows up in the force law for two particles of gravitating mass $m_G$ and $M_G$, where $M_G$ shall denote the earth's mass. $$ m_i \cdot \vec{a} = -G \cdot \frac{m_G \cdot M_G}{|\vec{r} - \vec{r} '|^2 } \cdot ...


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Here is an extremely simple explanation: Force = Mass x Acceleration Force / Mass = Acceleration Mass x Acceleration due to Gravity / Mass = Acceleration Acceleration due to Gravity = Acceleration For further intuition, consider this: The greater the mass, the greater the inertia. The greater the inertia, the greater the difficulty to accelerate the ...


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Without air resistance all objects are accelerated with gravity $$g = 9.81 \frac{m}{s^2}$$ Only the air causes a "slower" acceleration. This effect depends from the density and shape.


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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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