Gravitational fields question

Can you show your working out?

Reply 2

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.

v = \sqrt{\frac{GM}{R + h}}

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 = 4.2x10⁵kg (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

v = \sqrt{\frac{(6.7\text{x}10^{-11}\text{ x }6\text{x}10^{24})}{(6.4\text{x}10^6\text{ + }0.4\text{x}10^{6})}}

v = \sqrt{\frac{4.01\text{x}10^{14}}{6.8\text{x}10^6}}

v = 7689 ms^{-1}

rounds to

8kms^-1

(edited 5 years ago)

Reply 3

OP

Original post by uberteknik

Be careful with mixing up g, m, G and M.

v = \sqrt{\frac{GM}{r + h}}

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 = 4.2x10⁵kg (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

v = \sqrt{\frac{(6.7\text{x}10^{-11}\text{ x }6\text{x}10^{24})}{(6.4\text{x}10^6\text{ + }0.4\text{x}10^{6})}}

v = \sqrt{\frac{4.01\text{x}10^{14}}{6.8\text{x}10^6}}

v = 7689 ms^{-1}

rounds to

8kms^-1

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

Reply 4

Original post by Agent007

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

Use the same equation and set

F = mg

Then:

F = G\frac{Mm}{R^2}

mg = G\frac{Mm}{R_E^2}

R_E is the radius of the Earth

m cancels

g = G\frac{M}{R^2}

M = \frac{gR^2}{G}

M = \frac{9.81\text{ x }(6.4\text{x}10^6)^2}{6.75\text{x}10^{-11}}

M = 5.95\text{x}10^{24}\text{ kg}

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

v = \sqrt{\frac{G(\frac{gR^2}{G})}{r + h}}

G cancels leaving:

v = \sqrt{\frac{gR^2}{R + h}}

v = \sqrt{\frac{9.81\text{ x }(6.4\text{x}10^6)^2}{6.4\text{x}10^6\text{ + }0.4\text{x}10^6}}

v = \sqrt{\frac{4.02\text{x}10^{14}}{6.8\text{x}10^6}}

v = 7689\text{ ms}^{-1}

rounds to

v = 8 km s^-1

(edited 5 years ago)

Reply 5

Original post by Agent007

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

Are you now OK with my answers?

Reply 6

OP

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

Reply 7

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?

(edited 5 years ago)

Reply 8

OP

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.

Reply 9

Original post by Agent007

u acc genius. Thanks a lot.

Rep?

Spoiler

Reply 10

OP

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?

Reply 11

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

Reply 12

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?

Reply 13

OP

Original post by Isopher

I just did this

Did you understand it?

Reply 14

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.

Reply 15

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.

Reply 16

OP

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?

Reply 17

Original post by Agent007

But where did the 9x10^9 come from?

Thats the constant k that you stated 1/4piE0

Reply 18

OP