help with calculation problems (1 Viewer)

[abdooooo!!!]

1. Calculate the average acceleration required by the shuttle to achieve an orbital velocity of 7756 m.s-1, assuming that the rocket engines are running for 8.5 minutes. Express your answer in m.s-1 and as a multiple of g in the Earth's gravitational acceleration.

2. For the same shuttle launch in question 1, calculate:
a. The net force that must be exerted on an 80kg astronaut in the shuttle to achieve the same acceleration.
b. The apparent weight of the astronaut (or the force exerted by the shuttle seat on the astronaut). Express your answer in newtons and as a multiple of the astronaut's weight on Earth.

3. Calculate the velocity of a point P, on the surface of the Earth, near the equator, relative to the centre of the Earth. On a diagram indicate the direction of this velocity, given that the view of the Earth is form above the North Pole.

4. Calculate the orbital velocity of the Earth relative to the Sun.

5. The escape velocity for the Solar system from a point in the Earth's orbit is 42 kms-1. Use this information and the value for the orbital velocity of the Earth calculated in Q4, to identify the best time and position on Earth for the launch of an interplanetary probe that must leave the solar system. Mark this position on a diagram and then explain your choice.

please include working, thanks

 

[Constip8edSkunk]

1. assuming it is the average net acceleration they want, not the thrust...

let r = distance the rocket is from earths centre, a= avg net acceleration,

u = 0
v = 7756 ms^-1
t = 8.5*60 = 510 s

a = v-u / t, then divide by 9.8 to find how many g's

edit: Removed to clear confusion

2. let a = xg be the answer from 1, where x = some constant that i cant b buggered working out
a)
N = ma = 80*xg Newtons

b)
W = (x+1)mg Newtons



3. let R = radius of Earth at Equator (6700 km or sumthing)
v = sqrt[GM/R^2] (by equating centripetal and gravitational forces)
direction is tangential, anticlockwise when viewing from top from North pole



4. same as above with different value for R ...whatever the orbital radius is



5. Compare the velocity boost when shuttle is closest to the sun (launching opp. to the direction of spin, with direction or orbit) and when shuttle is furtherest from the sun (launching with the direction of spin and orbit) to see which 1 is more

 

[Ragerunner]

I thought calculations of orbital velocity isn't in the syllabus?

 

[Dash]

Originally posted by Ragerunner
I thought calculations of orbital velocity isn't in the syllabus?

I was thinking the same thing... meh

 

[abdooooo!!!]

Originally posted by Dash
I was thinking the same thing... meh

i wasn't thinking... ahhhh i can't understand any of the answers Constip8edSkunk posted

i guess i will just copy it and hope its all right

 

[Ragerunner]

I didn't even understand the question.

I absolutely hate maths in physics.

I much prefer the concepts. Thank god for the syllabus change

 

[Constip8edSkunk]

which part?
you work out the orbital velocity by equating gravitational and centripetal forces:

GMm/r^2 = mv^2/r
v^2=GM/r
v (orbital velocity) = sqrt[GM/r]

yeah they specific calculations regarding the orbital velocity is out of the syllabus , but i think they still can ask broad q about it. theres a dot point in section 3 of Space refering to quantitative and qualitative relationships of various properties(orbital radius, velocity, mass, period) of satellites and keplars law.... if i remember correctly.

 

[abdooooo!!!]

Originally posted by Constip8edSkunk
which part?
you work out the orbital velocity by equating gravitational and centripetal forces:

GMm/r^2 = mv^2/r
v^2=GM/r
v (orbital velocity) = sqrt[GM/r]

umm... i'm just now trying to comprehend your answers. i need some time to think...

 

[Ragerunner]

It's basically equating them and changing the subject so you get the velocity as the subject of the formula.

 

[abdooooo!!!]

yeah i get what you mean. but im really having major difficulties with the basics. see the answers, how many decimal place or significant value do you put?

 

[Xayma]

Originally posted by Ragerunner
I didn't even understand the question.

I absolutely hate maths in physics.

I much prefer the concepts. Thank god for the syllabus change

Bah it's people like you that made it change, grrr its your fault I wont do as well. Not to say I won't do as well, I just probably would of done better if it was more maths based.

 

[abdooooo!!!]

its people like me who benefits from such a fair system...

 

[zeropoint]

Originally posted by abdooooo!!!
yeah i get what you mean. but im really having major difficulties with the basics. see the answers, how many decimal place or significant value do you put?

When adding or subtracting quantities, the number of decimal places in the result should equal the number of decimal places in the quantity with the least number of decimal places. For example

1.12 + 2.1412 = 3.26

If the calculation involves multiplication or division, the number of _significant figures_ in the result should equal the number of significant figures in the least accurate quantity

1.120/2.1412 = 0.5231

 

[Ragerunner]

Originally posted by Xayma
Bah it's people like you that made it change, grrr its your fault I wont do as well. Not to say I won't do as well, I just probably would of done better if it was more maths based.

I didn't do anything!! *looks innocent*

When I picked Physics I thought it was Maths based. But at that time, my attitude towards the HSC was quite stupid. So I didn't care that I picked Physics/Chemistry.

Now I am so glad I did when they changed the syllabus.

 

[abdooooo!!!]

i think i figured most of it out, after 3 hours of looking at Constip8edSkunk and my text book. well this are my workings, please check it for me as this is an assignment due at the start of next week, thanks.

1.
a = v-u / t = 7756 ms^-1/510 s = 15.20... = 15 ms^-2

for multiple of g = a/g = 1.55... = 1.6

2.
a)
net force = ma = 80*(7756/510) = 1216.62... = 1.2*10^3 N

b)
apparent weight = ma + mg = 1216.62... + 80*9.8 = 2000.62... = 2.0*10^3 N

for multiple of g = apparent weight/normal true weight = 2000.62.../(80*9.8) = 2.55... = 2.6

3.
let R = 6.38*10^6 m (this is the figure for earth's radius, is this the same for radius at the equator)

v = sqrt[GM/r] = 7900.23... = 7.90*10^3 ms^-1

4.
let R = 1.50*10^11 m
let M = 1.99*10^30 kg

v = sqrt[GM/r] = 29747.04... = 2.97*10^4 ms^-1

5. i still don't get this one. whats going on? please explain to me or do the question for me. i need it done as soon as possible, its due next week

oh yea is my use of significant figures right?

 

[zeropoint]

Originally posted by abdooooo!!!
5. i still don't get this one. whats going on? please explain to me or do the question for me. i need it done as soon as possible, its due next week

oh yea is my use of significant figures right?

Velocity is a vector quantity, and as such can be added to other velocities by vector addition. An object `at rest' with respect to the surface of the Earth can be considered to possess two velocity vectors; the tangential velocity of the Earth with respect to the Sun v_ES as well as the surface velocity of the probe with respect to the Earth's centre v_PE, due to the rotation of the Earth about its axis. The maximum velocity will be realised when v_ES is parallel to v_PE. This occurs when probe lies on the equator pointing in the due easterly direction at around midnight. The velocity of the probe with respect to the Sun is thus

v_PS = v_PE + v_ES

 

[abdooooo!!!]

Originally posted by zeropoint
v_PS = v_PE + v_ES

hey what does these symbols exactly mean?

that is a pretty cool explaination/answer, thanks.

and can someone/anyone please check my workings... im desperate... are they right???

 

[zeropoint]

Originally posted by abdooooo!!!
hey what does these symbols exactly mean?

that is a pretty cool explaination/answer, thanks.

and can someone/anyone please check my workings... im desperate... are they right???

The equation v_PS = v_PE + v_ES is voiced; the velocity of the probe relative to the sun equals the velocity of the probe relative to the earth's centre of mass plus the velocity of the earth relative to the sun. By the way, I'm using _ to denote subscript and ^ to denote superscript, I'm not sure what the conventions of this board are.

For the first question, If orbital velocity is what you want, you can neglect the effect of gravity since I assume the rocket is to be fired horizontally from the appropriate alititude. In which case

a = Delta v / Delta t
= 7756 ms^-1 / 510 s
= 15.2 ms^-2
= 1.55 g

The net force experienced by the astronaut is simply

F = ma
= 80 kg * 15.2 ms^-2
= 1200 N

The apparent weight is just 1200 N. I'm starting to think you didn't really mean orbital velocity.

The velocity of a point on the equator is going to be

v = rw

where r is the radius of the Earth and w (really omega) is the angular velocity of the earth in rad/s.

v = 6.38*10^6 m * 2pi/86400 s^-1
= 464 ms^-1

To calculate the orbital velocity use

v = sqrt[GM/r]

where G is the universal gravitational constant, M is the mass of the sun and r is the radius of Earth's orbit

 

[abdooooo!!!]

Originally posted by zeropoint
The net force experienced by the astronaut is simply

F = ma
= 80 kg * 15.2 ms^-2
= 1200 N

The apparent weight is just 1200 N. I'm starting to think you didn't really mean orbital velocity.

yeah i did this... didn't i? i think you misread my answer or is it something else that i did wrong?

Originally posted by zeropoint
The velocity of a point on the equator is going to be

v = rw

where r is the radius of the Earth and w (really omega) is the angular velocity of the earth in rad/s.

v = 6.38*10^6 m * 2pi/86400 s^-1
= 464 ms^-1

oh yeah

but one thing i don't get. why does the v = sqrt[GM/r] only work for orbital velocity and not the rotational velocity, is it because the mass of the centre of the earth can't be obtained or something

cause i just used that equation with the mass of the whole earth... seems very wrong

this is all so confusing to me... thanks again

 

[zeropoint]

Originally posted by abdooooo!!!
yeah i did this... didn't i? i think you misread my answer or is it something else that i did wrong?

oh yeah

but one thing i don't get. why does the v = sqrt[GM/r] only work for orbital velocity and not the rotational velocity, is it because the mass of the centre of the earth can't be obtained or something

cause i just used that equation with the mass of the whole earth... seems very wrong

this is all so confusing to me... thanks again

You wrote that the apparent weight is 2000 N. It's not. Apparent weight is equal to the sum of the reaction forces on a body. In this case 1200 N. We ignore the mg term since it is perpendicular to the direction of motion, and as we well know from projectile motion, the force of gravity has no effect on the horizontal motion.

Imagine an arrow of length r extending from the centre of mass of the Earth to any point stationary with respect to the Earth's surface. The arrow will sweep out one complete revolution in 24 h regardless of its length. The tip of the arrow will experience a larger tangential velocity if the length of the arrow is increased. If the arrow turns through an angle @ in an interval of time t, it's angular velocity is

w = @/t

Note that w for the Earth is equal to 2pi radians (1 revolution) divided by 24*3600 s (the time taken for the Earth to complete 1 revolution). The tangential velocity of the arrow tip is

v = rw

Note also that an object orbiting the earth does not necessarily have an w equal to that of Earth. You can see this from Kepler's law of periods which states that

t^2 = k r^3

where k is a constant. Indeed the angular velocity of an object about a planet is in no way related to the angular velocity of the planet itself. Rather it is dependent on the mass of the planet and the radius of the orbit only. Orbital velocity is the tangential velocity v necessary to maintain an orbit of constant radius r.