Questions (1 Viewer)

[allGenreGamer]

Just two quick questions on physics:

1. According to Newton's Law of Universal Gravitation, a person standing next to a building should be attracted to it - why isn't this so?

2. An electron orbitsabout a nucleus in the Rutherford/Bohr model of the atom... what supplies the centripetal force in this scenario?

3. What's the difference between "The Slingshot effect" and 'Gravity assist"?

4. When the gravity assist effect is used,would an observer on that assisting planet agree that the probe' velocity has increased? Please explain.

Thanks

 

[Xayma]

1. They are, however this gravitational force is very small and the gravity towards the earth is relatively large. To give you an idea of how weak it is, two ships at dock are visibly attracted towards eachother, but you can overcome this with a very thin piece of rope.

2. In the Bohr model of an atom only the electromagnetic force between the positive and negative forces. This in turn leads to the electron spiraling into the atom very quickly.

 

[zeropoint]

Originally posted by allGenreGamer
need ask two more questions

3. What's the difference between "The Slingshot effect" and 'Gravity assist"?

The latter term is used by NASA.

4. When the gravity assist effect is used,would an observer on that assisting planet agree that the probe' velocity has increased? Please explain.

The gravitational field rotates the spacecraft's velocity in the planet frame leaving its magnitude unaffected. This is a speed change relative to the sun.

 

[Ragerunner]

1. There is a force of attraction, but it is negligible, using his law of universal graviation with the formula you can substitue values of one mass being very high and one mass being very low and get the answer.

 

[Rahul]

1. According to Newton's Law of Universal Gravitation, a person standing next to a building should be attracted to it - why isn't this so?
the gravity field between them wouldnt be strong enough to force the person onto the building. also, remember that this force would have to work against the earth's gravitational field.

2. An electron orbitsabout a nucleus in the Rutherford/Bohr model of the atom... what supplies the centripetal force in this scenario?
lol, no idea .
i am thinking that it would have to do with the forces between the negative electron and positive nucleus that would give it the centrifugal force. but then such motion would suggest acceleration, which in turn would spiral the electron to the nucleus.

3. What's the difference between "The Slingshot effect" and 'Gravity assist"?
nothing. they sound the same.

4. When the gravity assist effect is used,would an observer on that assisting planet agree that the probe' velocity has increased? Please explain
no real idea. :s
zeropoint, perhaps you could explain a bit more?

 

[allGenreGamer]

Thanks for the help i have yet another question...

A satellite of mass 1235kg is put into circular orbit, it circles the Earth exactly 9 times each day. Calculate th radius of the planet.

First, convert the period to seconds right? But then what ? I have the mass and period only, and I need to find the radius... wat formula(s) can I use? I don't know any that can help me find velocity,which will lead to the radius...

 

[Xayma]

The radius of the planet or radius of the orbit? Given that are you aren't given the altitude I would assume the radius of the orbit.

Ill show the working in a second.

We get to use lots of formulas
EDIT: I stuffed up the formulas, thats what I get for not looking at a formula sheet, Ill do it correctly soon.

Ok here it is again, stupid me I deleted the whole lot when I just needed to take out a factor:

F=mv²/r

Since it is in orbit F=mg

Cancelling out the m's we get g=v²/r

but g=6.67x10^-11x5.9742x10²⁴/r²

g=3.9848x10¹⁴/r²

Now velocity which in this case is actually speed =Circumfrence of orbit/T

ie g=(2(pi)r/T)²/r

g=4(pi)²r/T²

(The r's cancelled out)

3.9848x10¹⁴/r²=4(pi)²r/T²

Since we know T is 1/9 of a mean solar day we get

3.6724x10²²=4(pi)²r³

Making r the subject we get:

r=cuberoot(3.6724x10²²/4(pi)²)
r=9 761 789.54m
r= 9 761.78954km.

Much more logical then the wrong way's answer, and you dont even need the mass.

Waits for an error in calculation to show up. *Wonders how confusing it would be with r^2 instead of r² etc.


If you then wanted to calculate the radius of the planet you would need to know the gravity on the surface which is close enough to 9.8ms^-2 then plug it in g=GM₁/r²

but then you dont even need the satellites mass.

 

[allGenreGamer]

Yet another question regarding slingshot effect:
- In the slingshot effect, where does the extra energy come from?

 

[:: ck ::]

the gravitational force from the large mass of the planet

 

[Xayma]

The energy is the Gravititational Potential Energy of the object from the planet.

 

[:: ck ::]

so im wrong? 0.o

 

[Xayma]

No your not wrong, you just didn't answer the question. That is where the force is coming from but the energy used is the gravitational potential energy.

 

[:: ck ::]

ohh i get it.. haha thx

editz : actually .. didn i answer it? ... u answered wot type of energy it is... i answered where its from.. i think 0.o

where does the extra energy come from

 

[speersy]

quick question
rockets are fired from west to east right

 

[hainsay]

slingshot

b careful if asked to explain where the energy from the effect comes from. The energy comes from the orbital velocity of the planet. However, as we know, a large change in the ship's velocity will affect a negligible change the planet's velocity. The planet will eventually slow down and fall into the sun, but that's a problem for after a few million years

 

[kheir]

to speercy yes always towards east