Escape Velocity - Online Tutor, Practice Problems & Exam Prep

Escape Velocity

Hey, guys. So now that we've talked about velocities and gravitation, there's a very specific velocity that you're going to need to know called the escape velocity. Let's check it out. So basically, what the escape velocity is, it is the minimum launch speed that you need in order for an object to escape. And now what that word escape means, what does that mean? It means that this thing stops when it's very, very far away, and it can never come back towards the earth. So let's think about something for a second. So we're used to on earth, we'll throw an object upwards. It'll sort of reach the peak, and it'll come all the way back down because of the force of gravity. Right? We know that gravity pulls everything downwards. If you throw this thing a little bit harder, so we've got some initial right here if you throw this thing a little bit harder, it's still going to go up. Gravity is going to get a little bit weaker, but it's still eventually going to come crashing back down towards the Earth. The idea is that there is a launch speed. So there is a launch speed, a minimum launch speed that is so fast, so we're going to call that v_escape. Now what happens is it gets all the way up to this magically very, very far-away place. And what happens is that the force of gravity can't pull it back down towards the earth again. And so what happens is that it stops and never returns. Now we know, but when two objects get really far away from each other, the force of gravity approaches 0. So in other words, as this little r goes to infinity or gets really really really big, we know that the force of gravity is equal to g M m over r². So if this thing in the denominator gets really really big, then the force goes to 0. Okay? So what that means is that the object is going to stop, and when it stops very far away, the final velocity is equal to 0. So, in other words, v_final, when it gets all the way out here to this magical place that's infinitely far away, the final velocity is equal to 0. And the reason for that is that if it wasn't 0, if it were something that is greater than 0, then it wasn't the minimum launch speed that you could have thrown this with. So I'm going to make up a number here just for a second. Let's say you throw this up at, like, a 1000 meters per second, It gets up all the way out here, and the final velocity is equal to 5 meters per second. Well, then that means that you could have thrown it a little bit slower, and it still would have gotten out here with 0 meters per second. So that's what that means. You just have to throw it with the minimum launch speed. Alright? So this escape velocity here. And we know that we can't use kinematics because the accelerations are here, and we know that we can't use kinematics because the accelerations are constantly changing. So we have to use conservation of energy. So we've got initial kinetic and potential, plus any work done is equal to final kinetic and potential. Now, when you are throwing this object up with some initial velocity right here, we know that the kinetic energy is going to be 1/2 m v². So we've got math xmlns="http://www.w3.org/1998/Math/MathML"> 1 / 2 m v 2 , and now the initial gravitational potential energy is not 0, because you still have some distance away from the center of mass. So it's going to be negative g, big M, little m over little r. Now, we're talking about work. Sorry. We're talking about gravity. We know gravity is a conservative force. So that means there's no work done by nonconservative forces. It's kind of a double negative there. Now what about these last two? What about the energies when it finally reaches up this very, very far-away place? Well, we said that the energy or sorry. The velocity is equal to 0 when it finally gets out all the way over here. And we know that the kinetic energy is also 0 because it depends on this final velocity. Remember math xmlns="http://www.w3.org/1998/Math/MathML"> 1 / 2 m v 2 . So this thing is math xmlns="http://www.w3.org/1998/Math/MathML"> 1 / 2 m v final 2 , but we know that this thing is going to be equal to 0. So that means that the kinetic energy is going to be 0 there. Now what about the gravitational potential? Again, remember the equation for gravitational potential, it depends on that little r distance. So what happens is that as this distance gets really really big, then the gravitational potential energy will also go to 0. So that means that both of these things on the right will go to 0, and that allows us to figure out what the escape velocity is. So we're going to go a

a) How fast does a spaceship have to go to escape Earth, starting from the launch pad on the surface? Assume it burns all its fuel very fast and then shuts off the engines.

b) One idea to make getting to space easier is to build a space elevator, a large platform high above Earth's surface where spaceships can land and take off. How high above Earth would this platform have to be for the escape velocity to be 1/5 of its surface value? (For multiple choice, select the correct answer for part (b).)

• A. $2.52\times 10^3$ m
• B. $1.53\times 10^8$ m
• C. $1.59\times 10^8$ m
• D. $7.11\times 10^{10}$ m

Jumping Off an Asteroid

Hey, guys. Let's work out this problem together. So we're landing on the surface of a large spherical asteroid. We're going to set off walking in one direction, blah, blah, blah. We're going to drop some tools. And eventually, we have to figure out how fast do we have to jump in order to escape this asteroid. So what variable is that? What variable are we talking about here? We're talking about an escape velocity. So we can look at our escape velocity formula. So we've got v_escape is equal to 2GM/r. If I take a look at this equation, I need to figure out the escape velocity. I know that G is just a constant. I do not have the mass of the asteroid, and I also don't have the initial distance. I don't have any information about that. So this is a problem where we're going to have to basically go in a different direction and get what these variables are. We're going to have to go and get M and then go and get little r and then plug them back into this problem. I like this problem because it's going to combine a lot of different things from this chapter.

Let's first start out with just drawing a diagram. So I've got a large spherical asteroid that I am going to land on, like that. And the first part of this problem says that I'm going to set off walking in one direction, and after some later time, I've realized that I've basically come back around to, return to my spaceship. And I checked my little pedometer on my watch, and I've gone 25 kilometers. So that first part of the problem means that we are traveling around a spherical asteroid, which means that we're basically traveling around the circumference of it. This represents the circumference. And let's check out what that circumference means in terms of an equation. We know that the circumference of a circle is related to the radius of that circle by C=2πr. So we can actually go ahead and figure out what the radius of this circle is, this little r distance, by using this equation, C=2πr. We know what the circumference is. That's 25,000 in terms of meters. All we have to do is divide it by 2π, and we get the radius. And that's about 3,980 meters. So that is one part. So now we have the radius.

Now I need to go ahead and figure out what the mass is, and I'm going to do that over here somewhere. So I need to figure out what the mass of this asteroid is. So we're done with this first part of the problem. Now we've got to look at the second part. We're going to grab a tool from the toolbox and drop it, and then noting that, using our watch or something like that, it takes about 30 seconds to hit the ground from 1.4 meters high. So if we're looking for the big mass of the planet, we could use forces, gravitational forces, or gravitational accelerations. This part of the problem right here, this dropping 30 seconds and hitting the ground, looks like a kinematics problem, vertical kinematics, which we've done a lot of before. So I want to just go ahead and draw a quick diagram of what's going on in that part. So we've got this little, this ground right here, and I'm going to be dropping a rock, and I'm told that the distance that it falls is equal to 1.4. It takes a time of 30 seconds, and we know that it is dropped, which means that the initial velocity is equal to 0. So what is my target variable here? Well, I'm looking for, let's see. If I can keep going, this is going to be my Δy, right? It's going to be the distance this thing falls. I have the final velocity when it gets to the surface, but I don't know what that is. And I have the gravitational acceleration. That is going to be g at the surface. Right? Because this is going to be a gravitational acceleration, but we're standing really close to the surface of the asteroid. And I also don't know what that is. So if I take a look at my equations for M, my M equations involve forces and gravitational accelerations. Forces, I don't have any information about masses and forces, but I do have some sort of kinematics in which I have this acceleration. So I'm going to grab g_surface as the equation I'm going to use, because I'm on the surface, and that's going to be GM/r2. So now what happens is if I can figure out what this mass is, let me go ahead and rearrange for that. Right? So I've got g_surface,r2/G=M. So I just need to figure out what this g_surface is by looking at this part of the problem. So I basically got to go somewhere else to go and find out what g_surface is. Let's go ahead and do that. So if I'm looking for g_surface, I need that's a kinematic variable. I need 3 of the 5 other ones. Right? So I need, I've got the velocity initial, velocity final. I've got the t and the Δy. Which one of those do I which ones which one of these variables do I have? I have the initial velocity. I don't have the final velocity. I have the time and I have the Δy. So I'm going to pick the equation that doesn't have that v_final. Hopefully, you guys remember your kinematic formulas. This is going to be Δy equals initial velocity t+12gt2. Now, we know that the initial velocity is equal to 0, so that means that this is going to cancel out. And then we basically can choose this downward direction to be positive, because then everything turns out to be positive. So the Δy is going to be 1.4, that's the meters. And then we've got one-half of g, and this is going to be g_surface, and then the t is just equal to 30, and then we have that squared. So let me go ahead and move down. So basically, I'm going to move this one half over to the other side and then the 30 squared goes down. So we basically end up with 2 times 1.4 divided by 30 squared. Let me go ahead and reiterate that. Divided by 30 squared is equal to g_surface. So you go ahead and plug all that stuff into your calculator. You should get that g_surface=3.11×10-3. So now we have g_surface. So now we can plug this thing back into this equation. And if we go ahead and solve for M, we're going to get 3.11×10tothe-3,3,9802,dividedby6.67×10-11.G. And that's going to be the mass of the spherical asteroid, which is equal to We've got 7.39×1014. That's in kilograms. Now that I have this mass right here, I can plug this all the way back into that equation, and I can figure out what the final escape velocity is going to be. So it's basically just like working outwards to find all your variables and then plugging them all back into your original equation. 7×10tothe-11. Now I've got the mass of the asteroid, which is, 7.39×1014. So let me go ahead and actually move all of this stuff down. I'm going to move this stuff down a little bit. There we go. And then I've got, divided by r, the distance, which is going to be 3,980. You go ahead and plug all this stuff in. You're going to get the escape velocity is equal to 4.98. That's meters per second. So this is how fast you'd have to be jumping, which is actually probably doable if you were actually, like, in a spacesuit. So this is the final answer for how fast you'd have to be traveling to escape this asteroid. Let me know if you guys have any questions.

Here’s what students ask on this topic:

What is escape velocity and how is it calculated?

Escape velocity is the minimum speed required for an object to break free from a gravitational field without returning. It is derived using the conservation of energy principle, where the initial kinetic energy equals the gravitational potential energy at a distance where gravity approaches zero. The formula for escape velocity is:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝛂$ is the gravitational constant, $𝛱$ is the mass of the celestial body, and $𝑛$ is the distance from the center of the celestial body.

Why does escape velocity not depend on the mass of the object trying to escape?

Escape velocity does not depend on the mass of the object trying to escape because the mass cancels out during the derivation of the escape velocity formula. The formula is derived from the conservation of energy, where the initial kinetic energy and gravitational potential energy are set equal. Since both kinetic energy ($\frac{1}{2}𝑚{𝑛}^2$) and gravitational potential energy ($-𝛂𝛱𝑚𝑛∕𝑛$) include the mass of the object, it cancels out, leaving the escape velocity dependent only on the mass of the celestial body and the distance from its center.

How fast must an object be thrown to escape Earth's gravity?

To escape Earth's gravity, an object must be thrown with a speed of approximately 11.2 km/s. This speed is derived using the escape velocity formula:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝛂$ is the gravitational constant (6.67 × 10⁻¹¹ N(m/kg)²), $𝛱$ is the mass of Earth (5.97 × 10²⁴ kg), and $𝑛$ is the radius of Earth (6.37 × 10⁶ m).

What is the escape velocity from the Sun at Earth's orbital distance?

The escape velocity from the Sun at Earth's orbital distance is approximately 42.1 km/s. This is calculated using the escape velocity formula:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝛂$ is the gravitational constant (6.67 × 10⁻¹¹ N(m/kg)²), $𝛱$ is the mass of the Sun (2 × 10³⁰ kg), and $𝑛$ is the average distance from the Earth to the Sun (1.5 × 10¹¹ m).

How does the escape velocity change with distance from the celestial body?

Escape velocity decreases as the distance from the celestial body increases. This relationship is described by the escape velocity formula:

$\sqrt{\frac{2}{𝛂𝛱}}$

where $𝑛$ is the distance from the center of the celestial body. As $𝑛$ increases, the value under the square root decreases, leading to a lower escape velocity. This is because gravitational force weakens with distance, requiring less speed to overcome it.