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on Earth we have an escape velocity of about 11180 m/s and a gravitational field of g=9.81 m/s^2. It would seem obvious that the escape velocity was higher for stronger gravitational fields. Often it is. But Uranus has a gravity of 8.69 m/s^2, less than Earth, and its escape velocity is almost double (21300 m/s).

Looking at the formulas, the mathematics explains why, it depends on the radius: dependence on "r^2" for "g" and on "r" for "V" (gravitational potential, as we know used to calculate the escape velocity). So when the radius is very big compared to mass, it may lower "g" while "V" is higher.

But besides the mathematics (we can't simply say "the formulas state it is so!") how to explain in words (conceptually) why a weaker gravitational field (smaller "g") can have a higher gravitational potential and higher escape velocity? This seems paradoxical! Maybe this would be clear once one could figure out what the gravitational potential is. We find "g = nabla V". But still too much mathematics, I need an explanation in words.

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Imagine you had a small asteroid. The surface gravity isn't very strong because it's small and not very dense. If we could compress it, then we could approach closer to the center of mass and increase the surface gravity.

Let's pretend we take comet 67P ($1.0 \times 10^{13}\text{kg}$) and compress it until it is a sphere with a surface gravity equal to that at the earth's surface.

$$a = \frac{GM}{r^2}$$ $$r = \sqrt{\frac{GM}{a}}$$ $$r = 8.2\text{m}$$

It's impossibly tiny, but for now we can imagine landing on the surface and walking around almost as if we were on earth (because I'm very short and I'm ignoring tidal effects for a moment).

We get the high acceleration because we are so close to the mass. But what does that mean for the escape velocity? Now we take an earth launch vehicle and light it off. Because gravity at the surface is the same, the launch begins similarly.

But on earth, after the vehicle has gone up a 1km, the distance to the center of the planet has hardly changed. Gravity is still pulling very strongly and we continue to need a lot of fuel to keep going. This slow drop the gravitational field means that we have to do a lot of work to reach escape speed.

On our compressed comet though, rising 1km is already 2 orders of magnitude greater than our starting distance from the center. The gravitational field has already dropped to near zero. You can see that we've done much less work to reach escape speed.

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It is not just the value of the gravitational field strength $g$ which is important; the rate of change of gravitational field strength with distance from the centre of the planet $\dfrac{dg}{dr}$ needs to be considered.

Here are the graphs of gravitational field strength against height above surface for the Earth and Uranus.

enter image description here

The escape velocity from is dependent on the work done in taking a mass from the surface of a planet to infinity and the work done is the area under a graph force (per unit mass) vs distance graph.

By inspection the reason for the escape velocity of Uranus being greater than that of the Earth can clearly be seen.

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Two spheres with the same mass have the same gravitational field outside their boundaries. These spheres therefore also have the same escape velocity for projectiles starting at the same radius (the escape velocity being the velocity you need to ballistically escape to an infinite distance from some starting point).

If one of these two spheres is less dense, then it has a larger volume, and a larger radius, than the other. Therefore, if you are comparing the escape velocities on the surface of the two spheres, then the sphere with a larger radius clearly must have a smaller surface escape velocity, since a projectile starting at that larger radius has an advantage; it has already climbed out of a significant amount of the gravitational field before it started, and needs less velocity to escape the rest of the way.

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Consider an imaginary Uranus around earth and put the body on that surface and then see what will be value of g at that point for the two planets.

So, g is higher on the surface of earth because it has smaller radius but that imaginary surface onward, g for Uranus would be ~14 times that of earth because, its mass is ~14 times that of earth.

The velocity needed to rise up from earth's surface to the imaginary surface will counter balance some of the difference, but overall it is higher escape velocity for Uranus.

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But still too much mathematics, I need an explanation in words.

It's the mathematics, not the words, that makes physics today what it is. That said, I'll try to give an explanation in words. I will still use mathematics, however.

Suppose we have two planets, call them planets #1 and #2. Planet #1 has mass $M_1$ and radius $R_1$ while planet #2 has mass $M_2$ and radius $R_2$. Suppose that planet #1 is larger in size than planet #2: $R_1 > $R_2$. The goal is to find the conditions that enable the escape velocity from planet #1 to be greater than the escape velocity from planet #2, but the surface gravity on planet #1 is less than the surface gravity on planet #2.

In mathematics, this becomes $v_1 > v_2$ and $g_1 < g_2$, where $v=\sqrt{2\frac {GM}{R}}$ is the escape velocity an $g=\frac {GM}{R^2}$ is the surface gravity. It will help to use the average density, $M=\frac 4 3 \pi \rho R^3$. With a bit of elementary mathematics, the desired conditions become $$\begin{aligned}\rho_1 {R_1}^2 &> \rho_2 {R_2}^2\\ \rho_1 R_1 &< \rho_2 R_2\end{aligned}$$ This in turn leads to the desired result: $$\frac{R_1}{R_2} < \frac{\rho_2}{\rho_1} < \left(\frac{R_1}{R_2}\right)^2$$ As "less than" is a transitive relationship, one can also conclude that $\frac{\rho_2}{\rho_1} < \left(\frac{\rho_2}{\rho_1}\right)^2$ . This merely means that the smaller planet must necessarily be more dense than is the larger planet. That condition alone is not sufficient. We need to look at the two more restrictive inequalities to get the necessary conditions.

In words, those more restrictive inequalities say that ratio of the density of the smaller planet to that of the larger needs to be more than the ratio of the radius of the larger planet to the smaller one, but less than the square of this latter ratio. Both the escape velocity and surface gravity of the smaller planet will be less than those of the larger planet if the density ratio is less than the lower limit. Both will be greater than those of the larger planet if the density ratio exceeds the upper limit.

For example, consider a planet that is four times larger than another planet, in terms of radius. That's Uranus and the Earth. The desired conditions will be met if the smaller planet's density is at least 4 times but no more than 16 times that of the larger planet. The ratio of Earth's density to that of Uranus is 4.34, which is in this range.

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