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1

Whatever force you like. Since force just determines the rate change of velocity, you can use a massive force for a trillionth of a second or a tiny force for a long time period. However, if changed the velocity of the Earth (relative to the Sun) of 1 m/s, you'll would cause an impulse – change in momentum – on the Earth of $5.972×10^{24}$ kilogram-meters ...

0

Force is just a product of mass and acceleration and to calculate the force required to move the earth is quite easy. Though incredibly difficult to produce. IF we take the mass of earth as 5.9736 x 10^24 kg. And the acceleration 1 meter per second per second. And that is where you are wrong when you said 1 meter per second which is the unit for speed or ...

0

Your question makes sense if you replace "force" by "impulse". And that impulse is simply the mass of the Earth times $1 {\rm m\,s^{-1}}$. Moreover, you need to make the qualification: how much impulse is needed to change the Earth's motion state so that it is moving at $1 {\rm m\,s^{-1}}$ relative to its motion state now. According to the estimates of this ...

-5

I think everyone is wrong. It's dark matter that causes heaver metals to be pushed to the center and lighter metals/matter to "float" to the surface and beyond. Everything after the ozone is dark matter, an ocean of it! There are several factors to this... (1) The universe is expanding, and with it, dark matter. This causes an equally and accelerating ...

2

You have to compare the most probable speed of the atoms of He with the escape velocity from the planet. So you are working out whether $$\left(\frac{2kT}{m} \right)^{1/2} > \alpha\ \left(\frac{2GM}{R}\right)^{1/2},$$ where $T$ is the local temperature, $m$ is the He atom mass, $M$ is the planet mass and $R$ is the radius at which you are considering ...

0

It will oscillate around the center of Earth. To find the equation you will need to solve the generic oscillation problem: the gravity force / acceleration depends on the distance to Earth's center while everything which is further from the center will not impact on the acceleration. $g(x)=G\cdot M(x)/x^2$; \$M(x)=\mathrm{average~earth~density}\cdot ...

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