MedVision ad

Surfing Book- Space Topic Test Question (1 Viewer)

bleh1234

New Member
Joined
Feb 4, 2015
Messages
26
Gender
Undisclosed
HSC
N/A
33) A spacecraft is rising from the Earth's surface at 49m/s. At 980m the booster rocket tanks are jettisoned.
a) Calculate how long the booster tanks take to fall back to Earth.
b) Calculate the speed with which the booster tanks crash into the earth.
Really confused... I'm getting 15s and 147m/s but the answers are 20s and 245m/s.
Please explain if you know how to do it :(
Thanks!!!
 

Crisium

Pew Pew
Joined
Feb 17, 2014
Messages
2,009
Location
Australia
Gender
Male
HSC
2015
a) So when they say that the rockets are jettisoned they mean that they are shut down.

So assuming that it travels the same speed back down to Earth you do:

Time = Distance / Speed

Time = 980 / 49

Time = 20 seconds

b) Then you use the formula (delta y) = ut + 0.5at

You want to find out the vertical velocity here so you are looking for u (it should have a subscript of y)

The values you know:

(delta) y (otherwise known as the vertical displacement) = 980 m (Given in question)

t = 20 seconds (Solved above)

a (Acceleration due to gravity) = -9.8 (it is negative because it is acting downwards)

Sub these values into the equation

980 = 20u + (-9.8 x 20^2)

20u = 980 + 3920

20u = 4900

u = 245 ms^-1

I hope this helps :D
 

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
I do not agree with th above at all. (A) is assuming that the object is not accelerating and (b) the displacement should be -980, not +980

Use delta_y = uyt+a/2(t)^2

-980 = 49t -4.9t^2

Solve for t and it comes out as t=20s.

Then vy^2 =uy^2 +2aydelta_Y

Vy = -147 m/s


Moral of the story, don't trust the answers to the Surfing books
 
Last edited:

Crisium

Pew Pew
Joined
Feb 17, 2014
Messages
2,009
Location
Australia
Gender
Male
HSC
2015
I do not agree with th above at all. (A) is assuming that the object is not accelerating and (b) the displacement should be -980, not +980

Use delta_y = uyt+a/2(t)^2

-980 = 49t -4.9t^2

Solve for t and it comes out as t=20s.

Then vy^2 =uy^2 +2aydelta_Y

Vy = 147 m/s
Calm down I haven't done projectile in ages

EDIT:

You're argument for Part A is the same as what I said except I said that "assuming the speed stays the same" which means constant velocity which means no acceleration.
 
Last edited:

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
Calm down I haven't done projectile in ages
I'm very calm?

That's fine. I hope me pointing out some of the flaws in your method has helped you.

If it makes you feel any better, they must have been the same mistakes which the person who wrote the answers made :)
 

bleh1234

New Member
Joined
Feb 4, 2015
Messages
26
Gender
Undisclosed
HSC
N/A
I spent 30 minutes trying to do that question :'(
Thanks for all the help, I appreciate it :)
 

Crisium

Pew Pew
Joined
Feb 17, 2014
Messages
2,009
Location
Australia
Gender
Male
HSC
2015
I'm very calm?

That's fine. I hope me pointing out some of the flaws in your method has helped you.

If it makes you feel any better, they must have been the same mistakes which the person who wrote the answers made :)
...

It doesn't make me feel better, he's made so many mistakes in that book that I've lost count

+ For part A is it really necessary to use the delta_y formula in this scenario?
 

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
Calm down I haven't done projectile in ages

EDIT:

You're argument for Part A is the same as what I said except I said that "assuming the speed stays the same" which means constant velocity which means no acceleration.
Fair enough you made that assumption, it's just that the assumption is wrong. If this was a HSC question and you wrote that, you would get a big fat zero :)

Constant velocity, the boosters would continue to travel at 49m/s AWAY from the Earth and never reach the Earth :)
 

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
...

It doesn't make me feel better, he's made so many mistakes in that book that I've lost count

+ For part A is it really necessary to use the delta_y formula in this scenario?
Yes, you must as the object will accelerate due to gravity. It is only 980m above the surface of the Earth and we can assume that g=-9.8 at this Low altitude.

The way I solved it was a little different, I solved for Vy using the vy^2 eqn and then used v=u +at to solve for t, it's much easier to do on my phone calculator that way :)
 

Crisium

Pew Pew
Joined
Feb 17, 2014
Messages
2,009
Location
Australia
Gender
Male
HSC
2015
Fair enough you made that assumption, it's just that the assumption is wrong. If this was a HSC question and you wrote that, you would get a big fat zero :)

Constant velocity, the boosters would continue to travel at 49m/s AWAY from the Earth and never reach the Earth :)
It wasn't so I was generalising

But I would have written something along the lines of it falling back to the Earth at a constant velocity of 49 m/s
 

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
It wasn't so I was generalising

But I would have written something along the lines of it falling back to the Earth at a constant velocity of 49 m/s
I 100% know what you are saying, but I also know it's 100% incorrect to say it.

The fact that you got the correct answer is inconsequential, as in HSC Physics, we mark the PROCESS of obtaining the answer, not the answer.

When students make assumptions in order to answer questions, it just shows they have a lack of understanding on how to approach the question. There is no way you can assume that the object is in a state of uniform motion, it is a projectile which is subject to gravity.
 

bleh1234

New Member
Joined
Feb 4, 2015
Messages
26
Gender
Undisclosed
HSC
N/A
Hi again! :)
Sorry, i have another question.
The answers in the independent trial papers and the surfing book are different. The trial papers say that there is no net force acting on a satellite in stable orbit whereas surfing says that there is a net force caused by gravitational attraction towards the earth. Which one is correct? And can someone explain why? Both answers would seem to make sense to me...
 

Kaido

be.
Joined
Jul 7, 2014
Messages
798
Gender
Male
HSC
2015
Hi again! :)
Sorry, i have another question.
The answers in the independent trial papers and the surfing book are different. The trial papers say that there is no net force acting on a satellite in stable orbit whereas surfing says that there is a net force caused by gravitational attraction towards the earth. Which one is correct? And can someone explain why? Both answers would seem to make sense to me...
I've seen this question multiple times, and majority of trials have written no net force.
Now whether this is true for HSC exams, i'm not sure

However, from my understanding, the gravitational force is always acting on the satellite (hence keeping it in a stable orbit), hence the notion of 'centripetal force'
 

PhysicsMaths

Active Member
Joined
Dec 9, 2014
Messages
179
Gender
Male
HSC
2015
Hi again! :)
Sorry, i have another question.
The answers in the independent trial papers and the surfing book are different. The trial papers say that there is no net force acting on a satellite in stable orbit whereas surfing says that there is a net force caused by gravitational attraction towards the earth. Which one is correct? And can someone explain why? Both answers would seem to make sense to me...
Well here's my 2 cents..
I was just wondering, if there was a net force, then would it still be considered a stable orbit?
A net force would cause it to deviate from its circular path i think
 

Fizzy_Cyst

Owner @ Sigma Science + Phys Goat
Joined
Jan 14, 2011
Messages
1,212
Location
Parramatta, NSW
Gender
Male
HSC
2001
Uni Grad
2005
Well here's my 2 cents..
I was just wondering, if there was a net force, then would it still be considered a stable orbit?
A net force would cause it to deviate from its circular path i think
If there was no net force, the object would not change velocity.

The satellite is obvsly changing velocity as the direction of its motion changes.

There is a net force acting on the satellite, for a GEO, net force is gravity, for a LEO net force is vector sum of gravity + air resistance (collisions with atmosphere)

Which independent trial paper says there is no net force in a stable orbit?
 

bleh1234

New Member
Joined
Feb 4, 2015
Messages
26
Gender
Undisclosed
HSC
N/A
2006
2) An object is in stable orbit around the earth. Which statement about the force acting on this object is correct?
C) Gravitational force is the only force acting on the object
D) There is no net force acting on the object
D is the answer
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top