• Congratulations to the Class of 2024 on your results!
    Let us know how you went here
    Got a question about your uni preferences? Ask us here

Space Help (1 Viewer)

mrzeidan1

Tigers '08
Joined
Feb 17, 2006
Messages
201
Location
greenacre
Gender
Male
HSC
2007
Imagine that you were to travel in a spacecraft of mass 10,000kg from the Earth to the Moon. After liftoff, you intend to orbit the Earth at an altitude of 40,000km and then proceed to the moon.

I am asked in this question to investigate the acceleration due to gravity acting upon the spacecraft during its journey...

Do I need to take both the gravitational force of the Earth and the Moon into account?

If so, how can I work them out?

Thanks in advance
 

Heart.O.Gold

New Member
Joined
Nov 12, 2006
Messages
13
Gender
Male
HSC
2007
Imagine that you were to travel in a spacecraft of mass 10,000kg from the Earth to the Moon. After liftoff, you intend to orbit the Earth at an altitude of 40,000km and then proceed to the moon.

I am asked in this question to investigate the acceleration due to gravity acting upon the spacecraft during its journey...

Do I need to take both the gravitational force of the Earth and the Moon into account?

If so, how can I work them out?

Thanks in advance
I am guessing there are a few parts to this question;
---------------

At lift-off- the acceleration due to gravity is 9.8ms^-2
To work it out to equal 9.8ms^-2
g= G X (mass of Earth / Radius of Earth^2)

G= 6.67X10^-11 Mass of Earth = 5.97X10^24 Radius of Earth = 6.3X10^6)

------------------
Orbiting Altitude- acceleration due to gravity is 0.19ms^-2
g = G X (mass of Earth/ [radius of Earth + altitude]^2)

G= 6.67X10^-11, Mass of Earth = 5.97X10^24 Radius = 6.3X10^6 + 40 000 000

Therefore g = 0.19ms^-2
---------------
g at the moon - acceleration due to gravity is 1.6ms^-2
g= GX (mass of moon / Radius of moon^2)

G=6.67X10^-11 Mass of Moon = 7.35X10^22 Radius of moon = 1738
g=1.6ms^-2
----------------

Hope that is what you were looking for...
 

mrzeidan1

Tigers '08
Joined
Feb 17, 2006
Messages
201
Location
greenacre
Gender
Male
HSC
2007
Thanks mate. We also have to graph results taking 5 points into consideration. Does this mean that i need to calculate 2 points between the orbit and takeoff/landing?
 

Forbidden.

Banned
Joined
Feb 28, 2006
Messages
4,436
Location
Deep trenches of burning HELL
Gender
Male
HSC
2007
Heart.O.Gold said:
I am guessing there are a few parts to this question;
---------------

At lift-off- the acceleration due to gravity is 9.8ms^-2
To work it out to equal 9.8ms^-2
g= G X (mass of Earth / Radius of Earth^2)

G= 6.67X10^-11 Mass of Earth = 5.97X10^24 Radius of Earth = 6.3X10^6)

------------------
Orbiting Altitude- acceleration due to gravity is 0.19ms^-2
g = G X (mass of Earth/ [radius of Earth + altitude]^2)

G= 6.67X10^-11, Mass of Earth = 5.97X10^24 Radius = 6.3X10^6 + 40 000 000

Therefore g = 0.19ms^-2
---------------
g at the moon - acceleration due to gravity is 1.6ms^-2
g= GX (mass of moon / Radius of moon^2)

G=6.67X10^-11 Mass of Moon = 7.35X10^22 Radius of moon = 1738
g=1.6ms^-2
----------------

Hope that is what you were looking for...
Now I'll modify that a little and I'll use these tags to make it look easier ...

Code:
[sup][/sup]
Throughout all the working out, g will be the acceleration due to gravity in meters per second squared (ms-2) and G will be the universal gravitational constant (i.e 6.67x10-11 Nm2 kg-2)

At lift-off- the acceleration due to gravity is 9.8ms-2
To work it out to equal 9.8ms-2

me is the mass of Earth
re is the radius of Earth

Data:

g = ?
G = 6.67x10-11 Nm2 kg-2
Mass of Earth = me = 5.97x1024 kg
Radius of Earth = re = 6.3x106 m

g = Gme/re2



------------------

Hey, is this finding the acceleration due to gravity at 40,000km above the earth ?

Orbiting Altitude- acceleration due to gravity is 0.19ms-2

Data

Let ro be orbital radius of Earth
Orbital Radius of Earth = ro = re + ra
Where ra is the altitude


g = ?
G = 6.67x10-11 Nm2 kg-2
Mass of Earth = me = 5.97x1024 kg
Radius+Orbital Altitude of Earth = ro = 6.3x106 m + 40 000 000 m
(Don't forget your units.) (4.0x107 m)

g = Gme/ro2

Therefore g = 0.19ms-2

Therefore the higher the altitude the weaker the acceleration due to gravity, they are inversely proportional.
Sub in a big value for Orbital Altitude, you get a small acceleration
Sub in a small value for Orbital Altitude, you get a larger acceleration
Sub in no value for Orbital Altitude, you get the planet's original acceleration due to gravity on its surface (e.g on Earth's surface it's 9.8ms-2)

---------------

g at the moon - acceleration due to gravity is 1.6ms-2

G = 6.67x10-11 Nm2 kg-2
Mass of Moon = mm = 7.35x1022 kg
Radius of Moon = rm = 1738 m

g = Gmm/rm2

g =1.6ms2

I assume this is at the Moon's surface
---------------

By the way, which formula is the
g = Gm/r2 derived from ?

It has to do with escape velocity and centripedal acceleration ...
 
Last edited:

Heart.O.Gold

New Member
Joined
Nov 12, 2006
Messages
13
Gender
Male
HSC
2007
just got two questions.
1. how did you manage to use subsript and superscript?
2. Where the hell do u get off on remodeling my working weirdo?
 

airie

airie <3 avatars :)
Joined
Nov 4, 2005
Messages
1,143
Location
in my nest :)
Gender
Female
HSC
2007
Heart.O.Gold said:
just got two questions.
1. how did you manage to use subsript and superscript?
2. Where the hell do u get off on remodeling my working weirdo?
There's no need to call him a weirdo, he's just trying to grasp it o.0

Check out the link to vB codes under "Posting Rules" down the bottom. Just use the tags [sub ]-text-[/sub ] (without spaces) for subscripts and [sup ]-text-[/sup ] for superscript.
 

Heart.O.Gold

New Member
Joined
Nov 12, 2006
Messages
13
Gender
Male
HSC
2007
airie said:
There's no need to call him a weirdo, he's just trying to grasp it o.0

Check out the link to vB codes under "Posting Rules" down the bottom. Just use the tags [sub ]-text-[/sub ] (without spaces) for subscripts and [sup ]-text-[/sup ] for superscript.
Thanks for that;

I guess he isn't a weirdo, just taking my information and making it much more clearer.

I withdraw my comment
 

Forbidden.

Banned
Joined
Feb 28, 2006
Messages
4,436
Location
Deep trenches of burning HELL
Gender
Male
HSC
2007
Now I know where g = Gm/r2 is derived from ...

F = mg
F = Gm1m2/r2

So F = mg = Gm1m2/r2

Therefore:

g = Gm/r2

Cancel out the m from mg and remove either m1 or m2, does not matter which one then remove the subscript and thats it.

Heart.O.Gold said:
Thanks for that;

I guess he isn't a weirdo, just taking my information and making it much more clearer.

I withdraw my comment
Good.

airie said:
There's no need to call him a weirdo, he's just trying to grasp it o.0

Check out the link to vB codes under "Posting Rules" down the bottom. Just use the tags [sub ]-text-[/sub ] (without spaces) for subscripts and [sup ]-text-[/sup ] for superscript.
hi airie, yes i am grasping it, what I need to really grasp is projectile motion.
Projectiles are objects which move only under the influence of gravity, and can be projected in many different ways.
For example when an object is thrown into the air at time t, it will arrive back on the floor at time 2t, that is alone enough to throw me off track. Projectile motion is not all just "subbing in the values given" all the time, let alone not even knowing when to use the correct formula
 
Last edited:

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

Top