1) If the earth shrinks into a black hole, what will be the radius of the earth? Solution: Since the minimum escape speed is v(esp)=c=3x10^8m/s v(esp)=square root of (2GM/r) 3x10^8=square root of [2(6.67x10^-11)(5.98x10^24)/r)] r can be found. As we know, light CAN'T escape from a black hole. But, if we substitute the escape speed as the speed of light, that means speed can escape the black hole's gravitational field. Why can't we substitute c as the escape speed, as the solution does? 2) What is the total amount of energy needed to place a 2000kg satellite in circualr Earth orbit, at an altutide of 500km. [Is the answer simply Et=1/2 Eg=-5.80x10^10 J or is it the change in total energy or something? I don't get what the question is asking for.] 3) The mass of the Moon is 6.7x10^22 kg, and its radius is 1.6x10^6 m. If a woman can raise her centre of gravity 2.0m vertically in a high jump at earth's sruface. How high can she jump with the same muscular effort on the Moon's surface? [My question is, will the initial velocity of the woman be the same on the earth and on the moon? What I am thinking is that the earth and the moon have different values of Fg, so if she applies the same muscular strength (or same applied force), wouldn't the initial speed as she leaves the ground be different?] Thanks everyone for explaining! Please Register or Log in to view the hidden image!
Your exams approaching? You seem to have a lot of questions all of a sudden! There's a certain radius from the black hole where the escape velocity becomes the speed of light (I guess this means that the light can just escape here). This area's called the "event horizon." Once you pass it, there's no coming back - unless you call getting radiated out a particle at a time a "comeback." You're being asked to find the radius of the event horizon (also called the Swartzchild radius I think), so you just set the escape velocity as c. The amazing thing is that using Newtonian mechanics actually gets you the correct radius, even though you're technically using the wrong formulas. This is a general relativity problem, and usually Newton's old equations fail in such situations. You're being asked for an energy difference. The energy needed is the potential energy of the satellite in orbite relative to the surface of the Earth, plus the kinetic energy it needs if its going to stay in orbit. You can probably neglect the kinetic energy the satellite already has before launch, due to the rotation of the Earth. The energy difference sure as hell isn't negative. The initial velocities should be the same, but you don't need them. The potential energy difference in the jump stays the same.
Yes, exams are approaching. About 2 weeks away from now. Please Register or Log in to view the hidden image! 1) A black hole has the property that the escape speed is GREATER than the speed of light. (i.e. light can't escape a black hole) The solution did it this way, by substituting c as escape speed: 3x10^8=square root of [2(6.67x10^-11)(5.98x10^24)/r) This means that with a speed of 3x10^8m/s, light CAN escape a black hole, which is not true, right? So is the solution wrong? What is the escape speed at the event horizon? Is it greater than or equal to the speed of light? If the escape speed is equal to the speed of light, does it mean that light can actually escape a black hole? 2) I was told that Total energy = Eg + Ek, and I can find the total energy at the point where the satellite is in orbit (using the altitude and speed of the satellite). The questions asks for total energy, and I am not sure if this "Total energy" is the required answer... 3) Is it valid to say that the initial velocity will be the same on the earth and the moon and use the kinetic equations to solve this problem? But the weight is different, the applied force is the same, the time interval of upward acceleration (i.e. to jump) is the same, so wouldn't that initial velocity as she leaves the ground be completely different?
I'm right in the middle of mine Please Register or Log in to view the hidden image! The escape speed is c at the event horizon, so the solution gives the Swartzchild radius of the black hole. The mass of a black hole is usually considered to be a singularity - a point mass of infinite density. The escape velocity gets higher as you get closer to the singularity. The black hole itself is the volume inside the event horizon, where the escape velocity becomes greater than c. Here's a nice computer-generated image of a black hole I got off wikipedia that illustrates all this: Please Register or Log in to view the hidden image! They're asking for the total energy needed to get the satellite into orbit from the ground. They're asking for an energy difference. "Total" means that they want the sum ΔE<sub>g</sub> + ΔE<sub>k</sub>. You can model it this way, but its only true if the action of jumping off the ground is instantaneous. This isn't the easiest way to solve the problem, though. It'd be different, but the total work done during the jump would be the same. During the action of jumping off the ground, the woman gains more potential energy before even leaving the ground on Earth than she does on the moon, so you'd expect the initial velocity to be less. Anyway, same work done by woman ==> same final potential energy. Since ΔE<sub>g</sub> = wΔh (with the weight w = mg), you just have to find h on the moon knowing h on Earth and the different values of g.
Do universities have exams that late? I thought most have exams during April. But you still have time to answer questions in the middle of exams, that proves you are smart. Please Register or Log in to view the hidden image! 2) So basically calculate the change in gravitational potential energy (from ground to altutide of 500km) plus the kinetic energy of the satellite in orbit, am I right? If so, this is basically the work done needed to place a satellite at rest on earth to its orbit, right? 3) "It'd be different, but the total work done during the jump would be the same. " <---why the same? "...same final potential energy" <---once again, why? I don't quite get the idea.Please Register or Log in to view the hidden image!
I know that it's possible to start a course around January/February in the Netherlands and take the end of year exams around November/December, because I have a couple of friends doing it (I think this kind of academic year is the norm in the southern hemisphere, in fact). Different unis in different countries will have different systems. Or plain stupid. You have to watch it with sites like sciforums - they can get pretty addictive. Though I suppose my thinking skills have improved since I joined. It might help you to take part in some of the discussions here sometimes, instead of just post questions. Yep. As I said, there's a bit of an annoyance in that the satellite already has some KE due to the Earth's spin on it's own axis, but you can neglect this. What do you think would be the same about the 2 jumps? I'd expect the woman to apply the same force with her legs over the same distance, so same work.
3) "I'd expect the woman to apply the same force with her legs over the same distance" <---once she's in the air, there is no applied force, I think. Which distance do you mean? The thing I don't understand is why would the potential energy at max height on earth equal to that on the moon........ Please Register or Log in to view the hidden image!
The distance her centre of gravity moves up while she's still in contact with the ground. To jump, you crouch, and then push up your centre of gravity a certain distance before leaving the ground. The force and distance would be the same on the Earth as the moon. The athlete puts the same amount of energy into the jump on the moon as on the Earth. Gravity wouldn't affect her physical strength.
2) "Yep. As I said, there's a bit of an annoyance in that the satellite already has some KE due to the Earth's spin on it's own axis, but you can neglect this." The earth is rotating at a very high speed, which gives the object on the ground a very large kinetic energy. Why can we omit this large amount of kinetic energy in the calculations?
I have a couple more questions on this topic. I hope someone can help me. Thanks a lot! Please Register or Log in to view the hidden image! 4) A space shuttle ejects a 1.2x10^3 kg booster tank so that the tank is momentarily at rest, relative to Earth, at an altitude of 2.0x10^3 km. Neglect atmospheric effects. How much work is done on the booster tank by the force of gravity in returning it to Earth's surface? [The first thing I can think of is W=Fd(cos theta), but here the force of gravity is not constant so I can't use this formula. How can I find the work done by gravity, then?] 5a) Calculate the escape speed from the surface of the Sun: mass = 1.99x10^30 kg, radius = 6.96x10^8 m. 5b) What speed would an object leaving Earth need to escape from our solar system? [For 5a), I got 6.18x10^5 m/s, but I don't know how to proceed to part b)]
Yes. You can obtain the schwarzschild radius (ie the radius of the event horizon of a schwarzschild black hole) from the schwarzschild metric. If you're new to GR, the metric encodes space-time path information for objects in the appropriate field, eg the schwarzschild metric describes worldlines in the schwarzschild geometry (technically it computes the path length). You can work out the metric by solving the field equations, or you can find it online in hundreds (thousands?) of places. If you look at the metric in spherical coordinates (t,r,theta,phi) you can compute the event horizon radius by inspection. An observer at infinity in the schwarzschild geometry sees two singularities, ie areas where the metric becomes "singular" (for our purposes here singular = undefined). Have a look at the metric, shown in equation 20 at this page: -> http://scienceworld.wolfram.com/physics/SchwarzschildBlackHole.html. One of the singularities occurs when r=0, try plugging r into the factors leading dt^2 and dr^2 and see what happens. This is the central singularity. The other one occurs when the factor in brackets in front of dr^2 is undefined: what value of r makes that undefined and why? This second singularity is the event horizon - observers at infinity in the schwarzschild geometry see a singularity at the horizon (time dilation goes to infinity, speeds go to speed of light, etc). Now match your value for r to that obtained by the newtonian method - they agree! Please Register or Log in to view the hidden image! For those more comfortable with GR: what characteristics of the schwarzschild geometry account for this matching of results between GR and newton's theory?
6) Consider a geosynchronous satellite with an orbital period of 24h. What speed must the satellite reach during launch to attain the geosunchronous orbit? (Assume all fuel is burned in a short period. Neglect air resistance.) [On the ground, does the satellite have kinetic energy? Because for question 2, we do the calculations by the fact that the kinetic energy of the satellite when it's on the ground is zero. Should I do the same for this question?]
It'll just be the potential energy difference. I'd calculate the escape velocity from the surface of the Earth, given the Earth's distance from the sun (in other words, take the object's potential energy both in the Earth's and the sun's gravitational field into account).
The problem is that this initial kinetic energy depends on your location on Earth - it's zero at the poles, for example. Unless you're given extra information, you don't know what you need to know to take it into account. You could do some order-of-magnitude estimates to see if it's significant or not, though.
The question asks "...What speed must the satellite reach during launch to attain the geosunchronous orbit?" I am wondering how the launch (from the ground) alone can put the satellite in orbit. Is this possible? Wouldn't it need a rocket to re-direct the motion to put it in orbit around the earth?
4) Why only the potential energy difference? How about the kinetic energy difference? 5) How can I take both into account? I don't see what you mean...
Sorry - I was still thinking in terms of energy requirements. Good point. If you just catapult the satellite off the ground it'll try to follow an elliptical orbit that will pass through its launch position, so it'll crash back on Earth. Now the problem doesn't make enough sense to warrant a solution. The work the Earth does on the tank is the amount of potential energy it converts into kinetic energy. The object starts 6,400 km from the centre of the Earth, and 150 million km from the centre of the sun. It's initial PE is the sum of the -Gm/r for the Earth and for the sun.
