# Isaac Physics help

Watch
Announcements
#1
Does anyone know how to do this question? I have been stuck on it for ages.

Calculate the gravitational potential at a point 6.800×10^8m from the centre of the planet in question F5.16 on the same side of the planet as its moon. You should take the Universal gravitational constant to be G=6.674×10−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.]

(The question from F5.6 is: A 2.400×10^22kg moon orbits a 7.200×10^24kg planet with an orbital radius of 2.500×10^8m. Calculate the gravitational potential at the point half way between the centres of the planet and its moon. You should take the Universal gravitational constant to be G=6.674×10−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.] and the answer was -3.857x10^6 JKg^-1 )
0
2 years ago
#2
(Original post by laclare)
Does anyone know how to do this question? I have been stuck on it for ages.

Calculate the gravitational potential at a point 6.800×10^8m from the centre of the planet in question F5.16 on the same side of the planet as its moon. You should take the Universal gravitational constant to be G=6.674×10−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.]

(The question from F5.6 is: A 2.400×10^22kg moon orbits a 7.200×10^24kg planet with an orbital radius of 2.500×10^8m. Calculate the gravitational potential at the point half way between the centres of the planet and its moon. You should take the Universal gravitational constant to be G=6.674×10−11Nm^2kg^−2. [Note: more significant figures are used in this question than appear in earlier printings of the book.] and the answer was -3.857x10^6 JKg^-1 )

What are you stuck at?

If you can calculate gravitational potential at the point half way between the centres of the planet and its moon, I am not seeing why you cannot calculate the gravitational potential at a point 6.800 × 108 m from the centre of the planet in question F5.16 on the same side of the planet as its moon.

You are using the same principle.
0
1 year ago
#3
Did you ever get an answer to this? I got -7.067e^5 but it was wrong .
0
1 year ago
#4
(Original post by Ktweed)
Did you ever get an answer to this? I got -7.067e^5 but it was wrong .
It would be better that you show your working.
0
1 year ago
#5
(Original post by Eimmanuel)
It would be better that you show your working.
V=(-(6.674e-11)(7.2e24))/(6.8e8)
V=-7.067e5 JKg-1
0
1 year ago
#6
(Original post by Ktweed)
V=(-(6.674e-11)(7.2e24))/(6.8e8)
V=-7.067e5 JKg-1
0
1 year ago
#7
Yes the will definitely be a Gravitational Energy when the distance is the same order of magnitude away from the planet
0
1 year ago
#8
(Original post by Eimmanuel)
You need to add the gravitational potential due to the moon
What would r be to eork out thr gravitatipnal potential due to the moon though?
0
1 year ago
#9
(Original post by Ktweed)
What would r be to eork out thr gravitatipnal potential due to the moon though?
There are two distances - 2.500×10^8m and 6.800×10^8m.

Place the planet at the origin. The centre of the moon is located 2.500×10^8m from the origin.
From this, you should be able to work out the required "r".
0
1 year ago
#10
(Original post by Eimmanuel)
There are two distances - 2.500×10^8m and 6.800×10^8m.

Place the planet at the origin. The centre of the moon is located 2.500×10^8m from the origin.
From this, you should be able to work out the required "r".
Excellent, thank you, you are amazing! I've got the right answer now.
0
1 year ago
#11
If the orbital radius is 2.500e8, then surely the distance between the bodies is double that, so 5.000e8? I know it gives a wrong answer (using 2.5 as r instead of 1.25 gives you half the answer) but is it not a badly worded question?
0
1 year ago
#12
(Original post by tomhinde)
If the orbital radius is 2.500e8, then surely the distance between the bodies is double that, so 5.000e8?
Not sure how does orbital radius r translate to a separation of 2r.

(Original post by tomhinde)
I know it gives a wrong answer (using 2.5 as r instead of 1.25 gives you half the answer) but is it not a badly worded question?
From this statement, I have the impression that your understanding is incorrect.
1
1 year ago
#13
Don't worry, I'm stupid - I was thinking about circles having a diameter (which in this case would be the separation of the bodies), so radius would be half the separation. But when I started thinking about one planet and a satellite, I realised that I've got to take the planet to be the 'centre' of the 'circle', not outside the barycentre. Sorry for the dumb question.
0
X

new posts Back
to top
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

#### What do you want most from university virtual open days and online events?

I want to be able to watch in my own time rather than turn up live (189)
29.3%
I want to hear more about the specifics of the course (106)
16.43%
I want to be able to dip in and dip out of lots of different sessions (58)
8.99%
I want to meet current students (54)
8.37%
I want to meet academics and the people that will be teaching me (51)
7.91%
I want to have a taster lecture or workshop to see what the teaching is like (128)
19.84%
My parents/guardians are more interested than me to be honest (38)
5.89%
Other things – I'll tell you in the thread (21)
3.26%