13.3 Gravitational potential energy and total energy  (Page 3/6)

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$\frac{1}{2}m{v}_{\text{esc}}^{2}-\frac{GMm}{R}=\frac{1}{2}m{0}^{2}-\frac{GMm}{\infty }=0.$

Solving for the escape velocity,

${v}_{\text{esc}}=\sqrt{\frac{2GM}{R}}.$

Notice that m has canceled out of the equation. The escape velocity is the same for all objects, regardless of mass. Also, we are not restricted to the surface of the planet; R can be any starting point beyond the surface of the planet.

Escape from earth

What is the escape speed from the surface of Earth? Assume there is no energy loss from air resistance. Compare this to the escape speed from the Sun, starting from Earth’s orbit.

Strategy

We use [link] , clearly defining the values of R and M . To escape Earth, we need the mass and radius of Earth. For escaping the Sun, we need the mass of the Sun, and the orbital distance between Earth and the Sun.

Solution

Substituting the values for Earth’s mass and radius directly into [link] , we obtain

${v}_{\text{esc}}=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\left(6.67\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-11}\phantom{\rule{0.2em}{0ex}}\text{N}·{\text{m}}^{2}{\text{/kg}}^{2}\right)\left(5.96\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{24}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)}{6.37\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{6}\phantom{\rule{0.2em}{0ex}}\text{m}}}=1.12\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m/s.}$

That is about 11 km/s or 25,000 mph. To escape the Sun, starting from Earth’s orbit, we use $R={R}_{\text{ES}}=1.50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{11}\phantom{\rule{0.2em}{0ex}}\text{m}$ and ${M}_{\text{Sun}}=1.99\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{30}\phantom{\rule{0.2em}{0ex}}\text{kg}$ . The result is ${v}_{\text{esc}}=4.21\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{4}\phantom{\rule{0.2em}{0ex}}\text{m/s}$ or about 42 km/s.

Significance

The speed needed to escape the Sun (leave the solar system) is nearly four times the escape speed from Earth’s surface. But there is help in both cases. Earth is rotating, at a speed of nearly 1.7 km/s at the equator, and we can use that velocity to help escape, or to achieve orbit. For this reason, many commercial space companies maintain launch facilities near the equator. To escape the Sun, there is even more help. Earth revolves about the Sun at a speed of approximately 30 km/s. By launching in the direction that Earth is moving, we need only an additional 12 km/s. The use of gravitational assist from other planets, essentially a gravity slingshot technique, allows space probes to reach even greater speeds. In this slingshot technique, the vehicle approaches the planet and is accelerated by the planet’s gravitational attraction. It has its greatest speed at the closest point of approach, although it decelerates in equal measure as it moves away. But relative to the planet, the vehicle’s speed far before the approach, and long after, are the same. If the directions are chosen correctly, that can result in a significant increase (or decrease if needed) in the vehicle’s speed relative to the rest of the solar system.

Check Your Understanding If we send a probe out of the solar system starting from Earth’s surface, do we only have to escape the Sun?

The probe must overcome both the gravitational pull of Earth and the Sun. In the second calculation of our example, we found the speed necessary to escape the Sun from a distance of Earth’s orbit, not from Earth itself. The proper way to find this value is to start with the energy equation, [link] , in which you would include a potential energy term for both Earth and the Sun.

Energy and gravitationally bound objects

As stated previously, escape velocity can be defined as the initial velocity of an object that can escape the surface of a moon or planet. More generally, it is the speed at any position such that the total energy is zero. If the total energy is zero or greater, the object escapes. If the total energy is negative, the object cannot escape. Let’s see why that is the case.

when I click on the links in the topics noting shows. what should I do.
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SHREESH
thanks
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Given two vectors, vector C which is 3 units, and vector D which is 5 units. If the two vectors form an angle of 45o, determine C D and direction.
AFLAX
ty
Sharath
CD=5.83 n direction is NE
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state Hooke's law of elasticity
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Shaibu
thanks
Aarti
You are welcome
Shaibu
thnx
Junaid
what is drag force
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solve:A person who weighs 720N in air is lowered in to tank of water to about chin level .He sits in a harness of negligible mass suspended from a scale that reads his apparent weight .He then dumps himself under water submerging his body .If his weight while submerged is 34.3N. find his density
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The weight inside the tank is lesser due to the buoyancy force by the water displaced. Weight of water displaced = His weight outside - his weight inside tank = 720 - 34.3 = 685.7N Now, the density of water = 997kg/m³ (this is a known value) Volume of water displaced = Mass/Density (next com)
Sharath
density or relative density
Shaibu
density
Ian
Upthrust =720-34.3=685.7N mass of water displayed = 685.7/g vol of water displayed = 685.7/g/997 hence, density of man = 720/g / (685.7/g/997) =1046.6 kg/m3
1046.8
R.d=weight in air/upthrust in water =720/34.3=20.99 R.d=density of substance/density of water 20.99=x/1 x=20.99g/cm^3
Shaibu
Kg /cubic meters
Shaibu
Upthrust = 720-34.3=685.7N vol of water = 685.7/g/density of water = 685.7/g/997 so density of man = 720/g /(685.7/g/997) =1046.8 kg/m3
is there anyway i can see your calculations
Ian
Upthrust =720-34.3=685.7
Upthrust 720-34.3
=685.7N
Vol of water = 685.7/g/997
Hence density of man = 720/g / (685.7/g/997)
=1046.8 kg/m3
so the density of water is 997
Shaibu
Yes
Okay, thanks
Shaibu
try finding the volume then
Ian
Vol of man = vol of water displayed
I've done that; I got 0.0687m^3
Shaibu
okay i got it thanks
Ian
u welcome
Shaibu
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Same here, the function looks very ambiguous. please restate the question properly.
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Sharath
thank you so much Sharath Kumar
Liteboho
thank you, have also learned alot
Duncan
No issues at all. I love the subject and teaching it is fun. Cheers!
Sharath
cheers!
Liteboho
cheers too
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hii
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hii too
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haye
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shggggg
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prove using vector algebra that the diagonals of a rhombus perpendicular to each other.
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0
Keshav
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Keshav
o
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Sharath
Some corrections to my old post. Time taken to reach ground = 2*v*Sin (∆)/g R = (2*v²*Sin(∆)*Cos(∆))/g I put the g in the numerator by mistake in my old post. apologies for that. R on moon = (R on Earth)/(0.166)
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state Newton's first law of motion
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