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M3: Newtons law of gravitation watch

1. can anyone explain to me how I'd go about doing this question? thanks
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2. Well, first write down Newton's law of gravitation? You'll find some constants are unknown (e.g. mass of the earth), but ignore this for now.

Form equations for the acceleration at sea level, and the acceleration at 5km above sea level. You should be able to cancel out all the unknown quantities and be able to get the answer.
3. (Original post by DFranklin)
Well, first write down Newton's law of gravitation? You'll find some constants are unknown (e.g. mass of the earth), but ignore this for now.

Form equations for the acceleration at sea level, and the acceleration at 5km above sea level. You should be able to cancel out all the unknown quantities and be able to get the answer.
I was going to ask why do we have an equation at sea level but im guessing because we have information for that and we can form an equation that we can substitute later?
4. (Original post by DFranklin)
Well, first write down Newton's law of gravitation? You'll find some constants are unknown (e.g. mass of the earth), but ignore this for now.

Form equations for the acceleration at sea level, and the acceleration at 5km above sea level. You should be able to cancel out all the unknown quantities and be able to get the answer.
and here, why did he change the equation to a negative? is it because the rocket is traveling in the opposite direction to gravity?
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5. (Original post by Maths&physics)
I was going to ask why do we have an equation at sea level but im guessing because we have information for that and we can form an equation that we can substitute later?
im also having trouble with part b.

how did he get the equation: [( -1 / (r-x) ) + (1/r)] from part a?
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6. (Original post by Maths&physics)
I was going to ask why do we have an equation at sea level but im guessing because we have information for that and we can form an equation that we can substitute later?
Yes, exactly.

(Original post by Maths&physics)
and here, why did he change the equation to a negative? is it because the rocket is traveling in the opposite direction to gravity?
I don't see a negative anywhere here.

(Original post by Maths&physics)
im also having trouble with part b.

how did he get the equation: [( -1 / (r-x) ) + (1/r)] from part a?
Equation for potential energy (if you cover this in M3). Otherwise, you get the same result using "acceleration = v dv / dx" and integrating.
7. (Original post by DFranklin)
Yes, exactly.

I don't see a negative anywhere here.
sorry. I took the screenshot too early. so, why did he change the equation to a negative? is it because the rocket is traveling in the opposite direction to gravity?

Equation for potential energy (if you cover this in M3). Otherwise, you get the same result using "acceleration = v dv / dx" and integrating.
I get this bit now. thanks
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8. (Original post by Maths&physics)
sorry. I took the screenshot too early. so, why did he change the equation to a negative? is it because the rocket is traveling in the opposite direction to gravity?
Yes. are all positive quantities, but we see that acceleration is marked downwards, so we need .

Aren't these video tutorials?? Surely he would explain this step in it.
9. (Original post by RDKGames)
Yes. are all positive quantities, but we see that acceleration is marked downwards, so we need .

Aren't these video tutorials?? Surely he would explain this step in it.
he didn't explain it. hes usually quite good and thorough but not always.
10. (Original post by RDKGames)
Yes. are all positive quantities, but we see that acceleration is marked downwards, so we need .

Aren't these video tutorials?? Surely he would explain this step in it.

its fine. ive gone over my notes and found the way.
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11. (Original post by RDKGames)
Yes. are all positive quantities, but we see that acceleration is marked downwards, so we need .

Aren't these video tutorials?? Surely he would explain this step in it.
the escape velocity is the initial velocity?

but why are we ignoring v in the equation? what happens to it? thanks
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12. For escape velocity, v simply needs to be greater than 0 at infinite distance from the planet - so 0.5*v^2 > 0, so the RHS of the expression is greater than 0
13. And yes, the escape velocity is the initial velocity required for an object to 'escape' the gravitational field of the planet
14. (Original post by JSG29)
For escape velocity, v simply needs to be greater than 0 at infinite distance from the planet - so 0.5*v^2 > 0, so the RHS of the expression is greater than 0
but when I sub u back into the RHS of the equation, I get a negative value.
15. (Original post by Maths&physics)
can anyone explain to me how I'd go about doing this question? thanks
Gravitational attraction follows an inverse square law so the answer is just 9.8 * (6370/6375)^2.
16. (Original post by Maths&physics)
but when I sub u back into the RHS of the equation, I get a negative value.
How negative? If it's just a few thousand in this case, it's probably just a rounding error in the calculations made. If it's much bigger, you might have miscalculated - remember to convert radius into m from km, and use the u value in m/s
17. (Original post by JSG29)
How negative? If it's just a few thousand in this case, it's probably just a rounding error in the calculations made. If it's much bigger, you might have miscalculated - remember to convert radius into m from km, and use the u value in m/s
have you tried it using the values in the screenshoot?
18. (Original post by Maths&physics)
have you tried it using the values in the screenshoot?
Yes, using u=5036 I got 0.5v^2=-11683
Using u=5040, I got +8469
19. (Original post by JSG29)
Yes, using u=5036 I got 0.5v^2=-11683
Using u=5040, I got +8469
ah ok, so its rounding it that was the issue.

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