# A-Level Physics

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#1
ANY HELP WOULD BE APPRECIATED I'm stuck with Q4ii.

Q4. Communications satellites are usually placed in a geo-synchronous orbit.
(b) The mass of the Earth 6.00 × 1024 kg and its mean radius is 6.40 × 106m.
(i) Show that the radius of a geo-synchronous orbit must be 4.23 × 107 m.
(ii) Calculate the increase in potential energy of a satellite of 750 kg when it is raised from the Earth’s surface into a geo-synchronous orbit.

For Q4ii:
The solution is to do this:
V=(G)(6.0×10^24)(1/(6.0×10^6 )-1/(4.23×10^7 ))

Why can't you do:
V=(G)(6.0×10^24)/(6.40 × 10^6)-(4.23×10^7)

Also, Why is it (6.40 × 10^6)-(4.23×10^7) and not the other way round? = (4.23×10^7)-(6.40 × 10^6)

Many Thanks in advance to anyone who replies.
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4 weeks ago
#2
So the formula for gravitational potential,V is V = -(GM)/r. Remembering gravitational potential is negative, it is work against the field.
We think of gravitational potential as the energy required in moving an object from that point in the gravitational field to infinity, being 0 at infinity, which is why it is negative.
If we want to find the increase in gravitational potential, we do V_diff = -(GM)/r2 - (-(GM)/r1), remember to subtract the larger one (results in adding as of the double negative) because a difference is positive.
The question asks for the difference in potential between two points in the field, so at both points in the field, M is the same as it is the mass of the earth, and G of course is constant.
Thus we factor out the G and the M, giving V_diff = -GM(1/r2 - 1/r1)
So when you do V=(G)(6.0×10^24)/(6.40 × 10^6)-(4.23×10^7), you are subtracting a distance, r2 from gravitational potential, since they have different units, this doesnt work, you can't subtract a length from Energy per mass.

In reality the order of the two doesn't matter, you get the same answer just one is negative, you need the positive one as it asks for an increase in gravitational potential.
Last edited by johnthebaptist27; 4 weeks ago
1
#3
(Original post by johnthebaptist27)
So the formula for gravitational potential,V is V = -(GM)/r. Remembering gravitational potential is negative, it is work against the field.
We think of gravitational potential as the energy required in moving an object from that point in the gravitational field to infinity, being 0 at infinity, which is why it is negative.
If we want to find the increase in gravitational potential, we do V_diff = -(GM)/r2 - (-(GM)/r1), remember to subtract the larger one (results in adding as of the double negative) because a difference is positive.
The question asks for the difference in potential between two points in the field, so at both points in the field, M is the same as it is the mass of the earth, and G of course is constant.
Thus we factor out the G and the M, giving V_diff = -GM(1/r2 - 1/r1)
So when you do V=(G)(6.0×10^24)/(6.40 × 10^6)-(4.23×10^7), you are subtracting a distance, r2 from gravitational potential, since they have different units, this doesnt work, you can't subtract a length from Energy per mass.

In reality the order of the two doesn't matter, you get the same answer just one is negative, you need the positive one as it asks for an increase in gravitational potential.
"Thus we factor out the G and the M, giving V_diff = -GM(1/r2 - 1/r1)
So when you do V=(G)(6.0×10^24)/(6.40 × 10^6)-(4.23×10^7), you are subtracting a distance, r2 from gravitational potential, since they have different units, this doesnt work, you can't subtract a length from Energy per mass. "

Do you have to factorise it out though, since you have all the values and constants that you need?
So, if you plugged in the values, can you do V diff = -(GM)/r2 - (-(GM)/r1) without the factorising though?

Otherwise that makes a lot of sense now and I appreciate the extra info that you put in along side to help my understanding.
THANKS!! 0
4 weeks ago
#4
(Original post by 234UncleBob)
"Thus we factor out the G and the M, giving V_diff = -GM(1/r2 - 1/r1)
So when you do V=(G)(6.0×10^24)/(6.40 × 10^6)-(4.23×10^7), you are subtracting a distance, r2 from gravitational potential, since they have different units, this doesnt work, you can't subtract a length from Energy per mass. "

Do you have to factorise it out though, since you have all the values and constants that you need?
So, if you plugged in the values, can you do V diff = -(GM)/r2 - (-(GM)/r1) without the factorising though?

Otherwise that makes a lot of sense now and I appreciate the extra info that you put in along side to help my understanding.
THANKS!! No no, factorising is just for tidying up really , "V diff = -(GM)/r2 - (-(GM)/r1)" is exactly the same as "V_diff = -GM(1/r2 - 1/r1)", its just whatever you prefer, you wouldn't be marked down for not factorising it, just means you only have to type the G value and M value once into your calculator haha.
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#5
(Original post by johnthebaptist27)
No no, factorising is just for tidying up really , "V diff = -(GM)/r2 - (-(GM)/r1)" is exactly the same as "V_diff = -GM(1/r2 - 1/r1)", its just whatever you prefer, you wouldn't be marked down for not factorising it, just means you only have to type the G value and M value once into your calculator haha.
Haha, I'll remember that, Thanks.

Also probably the last thing before I've finished with the topic, what's the difference between the DEFINITIONS of Gravitational Potential and Gravitational Potential Energy?
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1 week ago
#6
(Original post by 234UncleBob)
Haha, I'll remember that, Thanks.

Also probably the last thing before I've finished with the topic, what's the difference between the DEFINITIONS of Gravitational Potential and Gravitational Potential Energy?
Do you have a physics text? I thought they are in the text.
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#7
(Original post by Eimmanuel)
Do you have a physics text? I thought they are in the text.
Definitions vary ad i know the examinations are looking for specific ones and i thought you would know them.
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