M3: Newtons law of gravitation

can anyone explain to me how I'd go about doing this question? thanks

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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.

Reply 2

Maths&physics

Original post by DFranklin

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?

Reply 3

Maths&physics

Original post by DFranklin

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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Reply 4

Maths&physics

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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Reply 5

DFranklin

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.

Reply 6

Maths&physics

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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Reply 7

RDKGames

Study Forum Helper

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.

G, M, m, d^2

are all positive quantities, but we see that acceleration is marked downwards, so we need

a<0

.

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

Reply 8

Maths&physics

Original post by RDKGames

Yes.

G, M, m, d^2

are all positive quantities, but we see that acceleration is marked downwards, so we need

a<0

.

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.

Reply 9

Maths&physics

Original post by RDKGames

Yes.

G, M, m, d^2

are all positive quantities, but we see that acceleration is marked downwards, so we need

a<0

.

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

this question follows the last. how would you suggest I go about this? thanks

its fine. ive gone over my notes and found the way. :smile:

(edited 6 years ago)

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Reply 10

Maths&physics

Original post by RDKGames

Yes.

G, M, m, d^2

are all positive quantities, but we see that acceleration is marked downwards, so we need

a<0

.

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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Reply 11

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

Reply 12

JSG29

And yes, the escape velocity is the initial velocity required for an object to 'escape' the gravitational field of the planet

Reply 13

Maths&physics

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. :frown:

Reply 14

SerBronn

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.

Reply 15

JSG29

Original post by Maths&physics

but when I sub u back into the RHS of the equation, I get a negative value. :frown:

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