# Gravitational fields question Watch

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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?

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?

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#3

(Original post by

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?

**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?

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 = 4.2x10

^{5 }kg (mass of the Space Station) and g = 9.81ms

^{-1}(gravitational acceleration)

The equation needs:

M (mass of the Earth) = 6x10

^{24}kg

and

G (gravitational constant) = 6.7x10

^{-11}Nm

^{2}kg

^{-2}

Yielding

rounds to

**8kms**

^{-1}

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reply

(Original post by

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

The equation needs:

M (mass of the Earth) = 6x10

and

G (gravitational constant) = 6.7x10

Yielding

rounds to

**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 = 4.2x10^{5 }kg (mass of the Space Station) and g = 9.81ms^{-1}(gravitational acceleration)The equation needs:

M (mass of the Earth) = 6x10

^{24}kgand

G (gravitational constant) = 6.7x10

^{-11}Nm^{2}kg^{-2}Yielding

rounds to

**8kms**^{-1}
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#5

(Original post by

what if they dont give you the mass of the earth?

**Agent007**)what if they dont give you the mass of the earth?

Then:

R

_{E}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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#6

(Original post by

what if they dont give you the mass of the earth?

**Agent007**)what if they dont give you the mass of the earth?

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(Original post by

Are you now OK with my answers?

**uberteknik**)Are you now OK with my answers?

thanks anyway

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#8

(Original post by

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

**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

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(Original post by

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?

**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?

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#10

(Original post by

u acc genius. Thanks a lot.

**Agent007**)u acc genius. Thanks a lot.

Spoiler:

Every little helps! (I'm close to 11 gems). Seriously sad, I know, I know. lol.

Show

Every little helps! (I'm close to 11 gems). Seriously sad, I know, I know. lol.

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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?

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?

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#12

**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?

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#13

(Original post by

I just did this

**Isopher**)I just did this

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#15

(Original post by

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?

**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?

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#16

(Original post by

Did you understand it?

**Agent007**)Did you understand it?

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(Original post by

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.

**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.

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#18

(Original post by

But where did the 9x10^9 come from?

**Agent007**)But where did the 9x10^9 come from?

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(Original post by

Thats the constant k that you stated 1/4piE0

**Isopher**)Thats the constant k that you stated 1/4piE0

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#20

**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?"

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?

_{0}) = 9 × 10^9

(½ × 5 × 10

^{−10}m)

^{2}= 6.25×10^-20 m

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