Q&A for How to Calculate Escape Velocity

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If I start "driving" away from earth at 100 mph and never speed up, won't I eventually get as far away as the rocket and therefore escape?

Community Answer

Yes. What is different, and what was not made clear perhaps, is that escape velocity is the velocity required at the start of movement for un-powered flight, as for example if you were to throw a stone. If you have enough fuel, then you can escape as slowly as you like. When I walk upstairs I am escaping the earths gravity a tiny bit, but I can't go on for long.

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I posit that a projectile requires constant propulsion to escape Earth's gravity. A friend maintains that a big enough "bang" at the Earth's surface can achieve the same result. Who is correct?

Community Answer

Both of you are. Conservation of energy does not depend on the propulsion that got you to space. Remember that escape velocity refers to the velocity of an object at sea level. If an explosion sends an object flying away at that speed, it will escape Earth. In your case, constant propulsion generates a constant force which steadily increases velocity, and is another (the practical) way to achieve escape velocity.

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Question

Does escape velocity vary on Earth?

Community Answer

In reality, yes, because our planet isn't a perfect sphere, which means the radius varies (lowest at poles & highest at equator). So the escape velocity varies from place to place. But for the sake of school physics problems and such, it's generally assumed that the escape velocity is the same all over the planet.

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If the weight of an object on the moon is 1/6 that on Earth, then what is the escape velocity of the moon?

Community Answer

Let's start from Newton's universal gravitational law, F = GMm/r^2. Weight is a measure of force, F = mg, where g is the gravitational acceleration. So, g = GM/r^2. We can therefore rewrite escape velocity as v = sqrt(2gr). On Earth, it would be g = 9.81 m/s^2. You are given that the weight of an object on the moon is 1/6th of that on Earth, so divide g by 6 and use the moon's radius to solve for v.

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What happens if the escape velocity is greater than the Earth's escape velocity sqrt? (2GM/R)

Community Answer

The object will have some velocity left after travelling an infinite distance from Earth.

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Is there a way to calculate escape velocity if the mass (m) of earth and gravitational acceleration (g) is given?

Community Answer

You would need the gravitational constant (G) to assist here... 6.67x10e-11. With the gravitational constant in play, and the mass of earth (m) known, you can back calculate the radius of the planet (g=Gm/r^2). With the radius of the planet known, you can then calculate escape velocity with v=(2GM/r)^.5.

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What must the present radius of the earth be compressed to so that the escape velocity is increased 10 times?

Community Answer

Escape velocity is inversely proportional to the square root of the radius: v ~ 1/sqrt(r). So If we want 10v, we have to decrease r by 100 (write 10v ~ 1/sqrt(r) and divide both sides by 10, and put the 10 in the square root). r_e = 6371 km, so r = 63.71 km.

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Can the escape velocity be negative?

Community Answer

"Velocity" means displacement per unit time. Displacement can only be >=0. So the answer is, No, escape velocity can never be negative.

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If the mass of the moon is 1/18 times that of Earth and its radius is 1/4 times the Earth's radius, and the escape velocity of Earth is 11.2 km/sec, what is its value at the surface of the moon?

Kal Mondal

Community Answer

ve = √(2GM/r) ve[moon] = √{2G(M/18)/(r/4)} = √{(2GM/r)*(4/18)} = 11.2•(√2/3) = 5.27973 km/s

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If I am calculating the escape velocity for Jupiter, will G still be equal to 6.67x10e-11?

Community Answer

Jupiter is a gaseous planet, therefore the equation would be 6.67x10e-6; the reason this occurs is because the gas giants are generally easier to get in and out of.

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What happens to a bullet that's fired into the air?

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It will come back to earth. The bullet cannot escape the earth's gravitational pull.

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Should r in Newton's Gravitational Force Equation be squared as gravitational force exponentially gets smaller as distance increases? If not, why?

Community Answer

Indeed, Newton's Gravity Equation reads: Fg = GMm/r^2. (Note that Fg is a force.) To derive the potential energy from this force, we need to integrate, and this is where the square disappears: U = integral from {r = infinite} to {r=r} of Fg dr. This equals -GMm/r @ {r=r} minus -GMm/r @ {r=infinite}, which boils down to U = -GMm/r.

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What does the "s" stand for in "11.2 km s-1"?

Community Answer

It stands for "speed," which is distance over time.

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Does the planet need constant density when calculating escape velocity?

Community Answer

Some parts of a planet are denser than others, but this variety in density is usually so small that it does not affect the result, so no, the density does not need to be constant.

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How do I calculate the escape velocity of a satellite launched from the earth's surface?

Community Answer

Use the equation v = sqrt(2 x G x M/r), where G is the gravitational constant, M is the mass of Earth and r is the radius of Earth.

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How can I calculate the escape velocity of a satellite launched from the Earth's surface if the radius of the Earth is 6.4×10^6?

Community Answer

v=(2GM/r)^1/2 v=((2*6.674*10^-11*5.98*10^24)/6.4*10^6)^1/2 v=11.139m/s

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