# Weight on a Neutron Star ?

1. Mar 24, 2005

### Logistics

Came across this forum while searching the net for help. It's great to have a forum dedicated to physics.

So hi to everyone :)

I'm stuck with this question.

If you weigh 660N on the earth, what would you weigh on the surface of a neutron star that has the same mass as our sun and a diameter of 22.0km ?

Take the mass of the sun to be m_s = 1.99×10^30kg, the gravitational constant to be G = 6.67×10^-11 N m^2/kg^2, and the acceleration due to gravity at the earth's surface to be g = 9.810 m/s^2.

My working out:

*m_n = mass of neutron
*m_me = mass of me

r_n = 11000m
m_n = 1.99 x 10^30kg
m_me = w/g = 660/9.810 = 67.27828746kg

= (G * m_n * m_me) / r_n^2

= (6.67x10^-11 * 1.99x10^30 * 67.27828746) / 11000^2

= 7.380205N

----

Now I'm told that the answer is wrong.

So can someone tell me where I went wrong ?

Last edited: Mar 25, 2005
2. Mar 24, 2005

### Andrew Mason

First of all, you can tell that your answer is wrong because you know the force of gravity is enormous on a neutron star and you have him weighing about 1/100th of his earth weight!

The problem is in your value for G. You forgot the minus sign in the exponent. That makes a difference of 22 orders of magnitude. The rest appears right.

AM

3. Mar 25, 2005

### Chronos

Andrew has it right. The measured weight increases exponentially [by the square] as you reduce the distance from the center of mass.

4. Mar 25, 2005

### Logistics

I forgot to write in the - but when I did the actual calculation I did have the minus.

ie. (6.67x10^-11 * 1.99x10^30 * 67.27828746) / 11000^2 =
7.380205N

And I'm told that's not the answer, and I think that the number should be a lot higher then 660N. not less than 660N.

Must be something else :(

Last edited: Mar 25, 2005
5. Mar 25, 2005

### tony873004

You could break it up into 2 steps. Compute your mass with
$$F=ma$$, re-write it to solve for m

For acceleration, either compute it with

$$a=\frac{GM_{Earth}}{r^2}$$

or just use
$$9.8 m/s^2$$
since this value is memorized by most students.

Then use the same formula for acceleration, but using the numbers for the neutron star. Then compute your Force on the neutron star.

6. Mar 25, 2005

### dextercioby

It can't be,check your multiplications & divisions again.

Daniel.

7. Mar 25, 2005

### Logistics

I went over it so many times grrrr :( Keep getting the wrong answer ie. 7.380205N

Could someone give it a try see what they get, please.

8. Mar 25, 2005

### dextercioby

Let $G_{S}$ be the weight of the astronaut on the surface of the neutron star.Applying Netwon's gravity law

$$G_{S}=G\frac{m_{astronaut}M_{star}}{R_{star}^{2}}\approx\frac{6.67\cdot 10^{-11}\cdot 67.28\cdot 1.99\cdot 10^{30}}{(11000)^{2}}\approx 7.38\cdot 10^{13}N$$

Daniel.

9. Mar 25, 2005

### Logistics

omfg I just realised why I kept getting told I got wrong answer.

I think it's time for me to go back to primary school.

7.38x10^13 isnt equal to 7.38.

What a dumb ass I am lol, can't even round of properly haha