Question

If you weigh 685 NN on the earth, what would be your weight on the surface of a neutron star that has the same mass as our sun and a diameter of 25.0 kmkm ? Take the mass of the sun to be msmsm_s = 1.99×1030 kgkg , the gravitational constant to be GGG = 6.67×10−11 N⋅m2/kg2N⋅m2/kg2 , and the free-fall acceleration at the earth's surface to be ggg = 9.8 m/s2m/s2 .

Answer 1

Answer:

W=5.92*10^13 N

Explanation:

The mass is

$F=Mg\\M=\frac{F}{g}=\frac{685N}{9.8\frac{m}{s^2}}=69.89kg$

Hence, the force in the nuetron star is

$F=G\frac{MM_n}{r^2}=(6.67*10^{-11}N\frac{m^2}{kg})\frac{(69.8kg)(1.99*10^{30}kg)}{(12.5*10^3m)^2}=5.92*10^{13}N$

Thus, it will be a great weigth!!

Hope this helps

Answer 2

Answer:

$W = 5.94 \cdot 10^{15} N$      

Explanation:  

To calculate the weight on the surface of a neutron star we can use the following equation:

$W = m*g$

Where:

W: is the weight of the person

m: is the mass of the person

g: is the gravity of the neutron star

Hence, first we need to find m and g. The mass is equal to:

$m = \frac{W}{g} = \frac{685 N}{9.81 m/s^{2}} = 69.83 kg$

Now, the gravity of the neutron star can be found using the followig equation:    

$F = \frac{G*m*M}{r^{2}} = m*g \rightarrow g = \frac{G*M}{r^{2}}$

Where:

G: is the gravitational constant = 6.67x10⁻¹¹ m³ kg⁻¹ s⁻²    

M: is the mass of the neutron star = 1.99x10³⁰ kg

r : is the distance between the person and the surface of the neutron star = 25/2 = 12.5 km  

$g = \frac{6.67 \cdot 10^{-11} m^{3}kg^{-1}s^{-2}*1.99 \cdot 10^{30} kg}{(12.5 \cdot 10^{3} m)^{2}} = 8.50 \cdot 10^{13} m/s^{2}$  

Now, we can find the weight on the surface of the neutron star:

$W = m*g = 69.83 kg * 8.50 \cdot 10^{13} m/s^{2} = 5.94 \cdot 10^{15} N$      

I hope it helps you!