# Gravitational fields questionWatch

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#1
Got this question from unified physics practice paper set 1

The ISS circles the Earth at a height 4x10^5 m. (h)
It's mass is 4.2x10^5 kg
The radius of the earth is 6.4x10^6 m (r)

Show that the speed of the ISS in orbit is 8 kms^-1

What i did was equate the centripetal force to the gravitational force:
mv^2/r = GMm/r^2 , the small m cancel out, one of the r's cancel out and rearrange to sqrt(GM/r+h)

I keep getting 2x10^-6

Can any1 help?
0
1 year ago
#2
Can you show your working out?
0
1 year ago
#3
(Original post by Agent007)
Got this question from unified physics practice paper set 1

The ISS circles the Earth at a height 4x10^5 m. (h)
It's mass is 4.2x10^5 kg
The radius of the earth is 6.4x10^6 m (r)

Show that the speed of the ISS in orbit is 8 kms^-1

What i did was equate the centripetal force to the gravitational force:
mv^2/r = GMm/r^2 , the small m cancel out, one of the r's cancel out and rearrange to sqrt(GM/r+h)

I keep getting 2x10^-6

Can any1 help?
Be careful with mixing up g, m, G and M.

is correct

The small m cancelled out (mass of the smaller object in orbit) and not the big M.

The capital G and capital M remain.

You incorrectly used m = 4.2x105 kg (mass of the Space Station) and g = 9.81ms-1 (gravitational acceleration)

The equation needs:

M (mass of the Earth) = 6x1024 kg

and

G (gravitational constant) = 6.7x10-11 Nm2kg-2

Yielding

rounds to

8kms-1
0
#4
(Original post by uberteknik)
Be careful with mixing up g, m, G and M.

is correct

The small m cancelled out (mass of the smaller object in orbit) and not the big M.

The capital G and capital M remain.

You incorrectly used m = 4.2x105 kg (mass of the Space Station) and g = 9.81ms-1 (gravitational acceleration)

The equation needs:

M (mass of the Earth) = 6x1024 kg

and

G (gravitational constant) = 6.7x10-11 Nm2kg-2

Yielding

rounds to

8kms-1
what if they dont give you the mass of the earth?
0
1 year ago
#5
(Original post by Agent007)
what if they dont give you the mass of the earth?
Use the same equation and set

Then:

RE is the radius of the Earth

m cancels

or, to make life easier, use the substitution for M (above) in the original expression for velocity:

G cancels leaving:

rounds to

v = 8 km s-1
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1 year ago
#6
(Original post by Agent007)
what if they dont give you the mass of the earth?
Are you now OK with my answers?
0
#7
(Original post by uberteknik)
Are you now OK with my answers?
yh im good. But still feels like a lot of work for a 4 mark question Calculating mass of earth, then subbing it in for the velocity of ISS seems a bit long.

thanks anyway
1
1 year ago
#8
(Original post by Agent007)
yh im good. But still feels like a lot of work for a 4 mark question Calculating mass of earth, then subbing it in for the velocity of ISS seems a bit long.

thanks anyway
I added an edit to post #5 and subbed the expression for M, which eliminated both M and m to give a much easier calculation using g and R+h. Have you seen that?
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#9
(Original post by uberteknik)
I added an edit to post #5 and subbed the expression for M, which eliminated both M and m to give a much easier calculation using g and R+h. Have you seen that?
u acc genius. Thanks a lot.
0
1 year ago
#10
(Original post by Agent007)
u acc genius. Thanks a lot.
Rep?

Spoiler:
Show

Every little helps! (I'm close to 11 gems). Seriously sad, I know, I know. lol.
3
#11
Can I ask another question? (My exam is tomorrow)

Hydrogen Iodide molecule consist of two ions 1H+ and 273-I (proton number of 53) held together by two electrostatic forces.
The charge of each ion has a magnitude of e. The ions are to be as point charges 5x10^-10 m apart. Calculate the magnitude of the resultant electric field strength E at the midpoint between the two ions.

I know resultant would be Resultant E = E(Iodine) - E(hydrogen) but they both have the same charge.
The mark scheme says:

E = kQ/r2 where k = 1/4πε0
E = 9 x 10^9 x 1.6 x 10^-19/6.25 x 10^-20
E = 2.3 x 10^10
2E = 4.6 x 10^10 (N C-1 )

Why did they even add. Is it because they are opposite charges?
Where did they get 9x10^9 and 6.25x10^-20 from?
0
1 year ago
#12
(Original post by Agent007)
Got this question from unified physics practice paper set 1

The ISS circles the Earth at a height 4x10^5 m. (h)
It's mass is 4.2x10^5 kg
The radius of the earth is 6.4x10^6 m (r)

Show that the speed of the ISS in orbit is 8 kms^-1

What i did was equate the centripetal force to the gravitational force:
mv^2/r = GMm/r^2 , the small m cancel out, one of the r's cancel out and rearrange to sqrt(GM/r+h)

I keep getting 2x10^-6

Can any1 help?
I just did this
0
1 year ago
#13
(Original post by Isopher)
I just did this
Has anyone done second part of this question on tatal energy i just couldnot do it? I can get the answer to whats in the Markscheme but i dont understand why?
0
#14
(Original post by Isopher)
I just did this
Did you understand it?
0
1 year ago
#15
(Original post by Agent007)
Can I ask another question? (My exam is tomorrow)

Hydrogen Iodide molecule consist of two ions 1H+ and 273-I (proton number of 53) held together by two electrostatic forces.
The charge of each ion has a magnitude of e. The ions are to be as point charges 5x10^-10 m apart. Calculate the magnitude of the resultant electric field strength E at the midpoint between the two ions.

I know resultant would be Resultant E = E(Iodine) - E(hydrogen) but they both have the same charge.
The mark scheme says:

E = kQ/r2 where k = 1/4πε0
E = 9 x 10^9 x 1.6 x 10^-19/6.25 x 10^-20
E = 2.3 x 10^10
2E = 4.6 x 10^10 (N C-1 )

Why did they even add. Is it because they are opposite charges?
Where did they get 9x10^9 and 6.25x10^-20 from?
I got the same answer as stated in the markscheme. Yes you do indeed add them together and you must always add when asked for resultant force/field between 2 like charges.
0
1 year ago
#16
(Original post by Agent007)
Did you understand it?
Yes the first part quite clearly but the second part on total energy is what I dont understand on this question.
0
#17
(Original post by Isopher)
I got the same answer as stated in the markscheme. Yes you do indeed add them together and you must always add when asked for resultant force/field between 2 like charges.
But where did the 9x10^9 come from?
0
1 year ago
#18
(Original post by Agent007)
But where did the 9x10^9 come from?
Thats the constant k that you stated 1/4piE0
0
#19
(Original post by Isopher)
Thats the constant k that you stated 1/4piE0
oh yh thats dumb of me. How about the 6.25x10^-20 come about?
0
1 year ago
#20
(Original post by Agent007)
Can I ask another question? (My exam is tomorrow)

Hydrogen Iodide molecule consist of two ions 1H+ and 273-I (proton number of 53) held together by two electrostatic forces.
The charge of each ion has a magnitude of e. The ions are to be as point charges 5x10^-10 m apart. Calculate the magnitude of the resultant electric field strength E at the midpoint between the two ions.

I know resultant would be Resultant E = E(Iodine) - E(hydrogen) but they both have the same charge.
The mark scheme says:

E = kQ/r2 where k = 1/4πε0
E = 9 x 10^9 x 1.6 x 10^-19/6.25 x 10^-20
E = 2.3 x 10^10
2E = 4.6 x 10^10 (N C-1 )

Why did they even add. Is it because they are opposite charges?
Where did they get 9x10^9 and 6.25x10^-20 from?

"Why did they even add. Is it because they are opposite charges?"
Yes and no.
You need to determine the direction of electric field at the midpoint due to the positive and negative charges.

Where did they get 9×10^9 and 6.25×10^-20 from?
k = 1/(4πε0) = 9 × 10^9

(½ × 5 × 10−10 m)2 = 6.25×10^-20 m
0
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