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Calculate the mass of the sun

Question:The radius of the Earth’s orbit about the Sun is 1.50 × 1011 m.Calculate the mass of the Sun.

Previous part of the question got you to to get the equation: 4pi^2/GMsun=T^2/r^3

T = 1 year = 3.16 × 107s [1]
ms = 4π2r 3/GT2 = [4π2 × (1.50 × 1011)3]/
[6.67 × 10–11 × (3.16 × 107) Rest of subs [1]
Mass of Sun = 2.00 × 1030kg [1] [3] 12

I can see how they got the answer. However initially i didn't get the previous part. So I attempted to answer the equation using the formula:
gravity=GMsun/r^2

so 9.81=6.67*10^-11(Msun)/(1.5*10^11)^2 giving an anser of 3.31*10^33kg. Could anyone explain why the second method is incorrect.
Original post by Tobeadoc
Question:The radius of the Earth’s orbit about the Sun is 1.50 × 1011 m.Calculate the mass of the Sun.

Previous part of the question got you to to get the equation: 4pi^2/GMsun=T^2/r^3

T = 1 year = 3.16 × 107s [1]
ms = 4π2r 3/GT2 = [4π2 × (1.50 × 1011)3]/
[6.67 × 10–11 × (3.16 × 107) Rest of subs [1]
Mass of Sun = 2.00 × 1030kg [1] [3] 12

I can see how they got the answer. However initially i didn't get the previous part. So I attempted to answer the equation using the formula:
gravity=GMsun/r^2

so 9.81=6.67*10^-11(Msun)/(1.5*10^11)^2 giving an anser of 3.31*10^33kg. Could anyone explain why the second method is incorrect.


That equation gives a value for "g" (gravitational field strength) at the surface of a spherical mass of radius r.
The value of g you used is the value on the surface the Earth. To use that equation it would need to be the value of "g" on the surface of the Sun. It certainly isn't 9.8 m/s/s
Secondly the radius used in that formula is the radius of the spherical mass. So it would have to be in this case the radius of the Sun.
So it's incorrect on 2 counts.
Reply 2
Original post by Stonebridge
That equation gives a value for "g" (gravitational field strength) at the surface of a spherical mass of radius r.
The value of g you used is the value on the surface the Earth. To use that equation it would need to be the value of "g" on the surface of the Sun. It certainly isn't 9.8 m/s/s
Secondly the radius used in that formula is the radius of the spherical mass. So it would have to be in this case the radius of the Sun.
So it's incorrect on 2 counts.

Thanks soo much for the answer. for the quantity e^-work function/temperature. how is there no units?
Original post by Tobeadoc
Thanks soo much for the answer. for the quantity e^-work function/temperature. how is there no units?


If you mean e raised to a power then
Any pure number (and e is certainly one) raised to a power is still a pure number with no units.
Pure numbers like e and Pi have no units.

Where is the expression e raised to power work function / temperature from?
Reply 4
Original post by Stonebridge
If you mean e raised to a power then
Any pure number (and e is certainly one) raised to a power is still a pure number with no units.
Pure numbers like e and Pi have no units.

Where is the expression e raised to power work function / temperature from?




This emission
is called the thermionic emission current, and the current per unit area
of the metal is the thermionic emission current density. The equation
giving the thermionic emission current density J at a kelvin
temperature T is J = A0T 2e^– φ/kT where A0 is a constant, φ is the work function and k is the Boltzmann
constant. To obtain the work function, the current density J is measured
at a number of temperatures T. Its at the end of a paper, designed to test you application skills.
Original post by Tobeadoc
This emission
is called the thermionic emission current, and the current per unit area
of the metal is the thermionic emission current density. The equation
giving the thermionic emission current density J at a kelvin
temperature T is J = A0T 2e^– φ/kT where A0 is a constant, φ is the work function and k is the Boltzmann
constant. To obtain the work function, the current density J is measured
at a number of temperatures T. Its at the end of a paper, designed to test you application skills.


I asked because in your post above you had

eWTe ^ {\frac{-W}{T}}

which cannot be correct.
This is because the power term must also be dimensionless. IE have no units, because you can only raise a number to a pure number.
Work function / temperature is not dimensionless.

However you now have the correct formula which has the power term as
eWkTe ^ {\frac{-W}{kT}}

The inclusion of the Boltzmann constant k now ensures that the power term is also dimensionless.
Reply 6
Original post by Stonebridge
I asked because in your post above you had

eWTe ^ {\frac{-W}{T}}

which cannot be correct.
This is because the power term must also be dimensionless. IE have no units, because you can only raise a number to a pure number.
Work function / temperature is not dimensionless.

However you now have the correct formula which has the power term as
eWkTe ^ {\frac{-W}{kT}}

The inclusion of the Boltzmann constant k now ensures that the power term is also dimensionless.


Ah i get it now...left out the units for the Boltzmann constant which now makes it understandable.

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