# Escape Velocity depends on the Radius of the Earth?

In summary: They're delimiters for LaTeX. The stuff between them ought to be interpreted by a Javascript package called MathJax abd shown to you as nicely rendered maths. You shouldn't need to do anything for this to happen - the maths should just render a second or two after the page loads.
We have 2 different formulas for escape velocity.
and
. If we look at the first formula we see that escape velocity is inversely proportional to the square root of Radius of Earth. While in the second formula, escape velocity is directly proportional to the square root of Radius of Earth.

We know that escape velocity depends on the Radius of Earth. Which one will show its proper dependence? Is it directly proportional or inversely proportional?

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Does ##g## depend on ##R##?

They're basically the same formula; how do you write ##g## in terms of ##G##, ##M## and ##R##?

But to answer your question, to find the dependence of one quantity (##V_{esc}##) on another (##R##), you have to write a relationship between them using only other quantities which are independent of those two things.

It's up to you to tighten the parameters of the question a little bit: i.e. is the density of the planet assumed constant?

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Ibix said:
Does ##g## depend on ##R##?
Yes. It is inversely proportional to square of radius. So if radius would increase, g would decrease and as a result V will decrease. While on the other hand if radius increases, V decreases. So ultimately V should be inversely proportional to the square root of R. Right?

Yes. It is inversely proportional to square of radius. So if radius would increase, g would decrease and as a result V will decrease. While on the other hand if radius increases, V decreases. So ultimately V should be inversely proportional to the square root of R. Right?

Use some algebruh! You might also say that ##M(R) = \frac{4}{3}\pi R^3 \rho##... or if the density is not uniform it gets more complicated...

etotheipi said:
Use some algebruh! You might also say that ##M(R) = \frac{4}{3}\pi R^3 \rho##... or if the density is not uniform it gets more complicated...
Thank you so much! It was so confusing to me. My doubts are clear now. Thanks again!

Thank you so much! It was so confusing to me. My doubts are clear now. Thanks again!

This is a common theme to look out for. For example, for uniform circular motion we have:
$$a = \frac{v^2}{R} = \omega^2 R$$
So, if you ask "how does acceleration depend on radius?", then it depends what other factors are held constant:

If velocity is constant, then acceleration is inversely proportional to radius;
If angular velocity is constant, then acceleration is proportional to radius.

You always have to ask yourself what quantities are fixed and what quantities are varying.

sophiecentaur, Nugatory, Delta2 and 3 others
Or to put it another way, the Escape Velocity from the surface of a planet always equals he velocity (neglecting air resistance) at which a body would impact the surface after falling from an altitude equal to the planet's radius at a constant acceleration equal to the planet's surface gravity.

By a well-known theorem, if V = Escape velocity (ie the final velocity of an object falling from infinity),
R = the radius of the planet and G = acceleration due to gravity, then V = sqrt(2GR)

But reverting to grammar school physics, if a body is falling under a *constant* acceleration G then V= Gt where t = time taken to fall, and D = (Gt^2)/2 where D = the distance of fall.

So if sqrt(2GR) = Gt then 2GR = (Gt)^2 and R = (Gt^2)/2

Hence D = R.

So escape velocity is determined by the surface gravity and the radius.

BvU
You can also argue with the energy. With the usual choice of the gravitational potential (going to 0 at infinity), $$\Phi=-\frac{G m_1 m_2}{r}$$
The minimal energy for an unbound orbit is ##E=0##
$$\frac{m_2}{2}v^2-\frac{G m_1 m_2}{r}=0,$$
with the radius of the Earth, ##R##
$$v=\sqrt{\frac{2 G m_2}{R}}.$$
Now the gravitational acceleration on the surface of the Earth is
$$g=\frac{G m_2}{R^2}$$
and thus
$$v=\sqrt{2gR}.$$

etotheipi
Or, the potential difference between the surface and infinity is:
$$V = \frac{GM}{R}$$
And, in the constant gravity model, the potential at a height ##h## is given by:
$$V = gh = \frac{GM}{R^2}h$$
These are equal when ##h = R##.

etotheipi
PeroK said:
$$V = \frac{GM}{R}$$
And, in the constant gravity model, the potential at a height ##h## is given by:
$$V = gh = \frac{GM}{R^2}h$$
These are equal when ##h = R##.

There seem to be a lot of hash and dollar signs in some posts. Are they mathematical symbols or is my computer rendering them wrongly?

Mikestone said:
There seem to be a lot of hash and dollar signs in some posts. Are they mathematical symbols or is my computer rendering them wrongly?

That's Latex. You should see them in mathematical form. Not sure why there would be a problem. You don't need anything special in your browser.

Mikestone said:
Are they mathematical symbols or is my computer rendering them wrongly?
They're delimiters for LaTeX. The stuff between them ought to be interpreted by a Javascript package called MathJax abd shown to you as nicely rendered maths. You shouldn't need to do anything for this to happen - the maths should just render a second or two after the page loads. So $$F=\frac{GMm}{r^2}$$should appear as Newton's law of gravity. Try the usual stuff (refresh, close your browser and reopen) if you don't see it - it's really important for this site...

Thanks. Things seem to be working ok on Google Chrome.

## 1. What is escape velocity?

Escape velocity is the minimum speed that an object needs to achieve in order to escape the gravitational pull of a celestial body, such as the Earth.

## 2. How does the radius of the Earth affect escape velocity?

The larger the radius of the Earth, the lower the escape velocity will be. This is because the gravitational pull of the Earth decreases with distance, so a larger radius means that the object is farther away from the center of the Earth and therefore experiences less gravitational pull.

## 3. How is escape velocity calculated?

The formula for calculating escape velocity is V = √(2GM/R), where V is the escape velocity, G is the gravitational constant, M is the mass of the celestial body, and R is the radius of the celestial body.

## 4. Why is escape velocity important?

Escape velocity is important because it determines whether or not an object can break free from the gravitational pull of a celestial body. It is crucial for space exploration and launching spacecraft into orbit.

## 5. How does escape velocity change on other planets?

The escape velocity on other planets depends on their mass and radius. Generally, planets with larger masses and smaller radii will have higher escape velocities, while those with smaller masses and larger radii will have lower escape velocities. For example, the escape velocity on Mars is about half of that on Earth due to its smaller size and mass.

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