To develop this problem it is necessary to apply the concepts related to the proportion of a neutron star referring to the sun and density as a function of mass and volume.
Mathematically it can be expressed as
Where
m = Mass (Neutron at this case)
V = Volume
The mass of the neutron star is 1.4times to that of the mass of the sun
The volume of a sphere is determined by the equation
Replacing at the equation we have that
Therefore the density of a neutron star is 
Answer:
25m/s²
Explanation:
Using one of the equations of motion.
v² = u²+2as where
v is the final velocity of the astronaut
u is his initial velocity
a = -g (the acceleration will be acceleration due to gravity since he is acting under the influence of gravity. The value is negative because the astronaut jumps up to a particular height)
s = H = total height covered
The equation will then become;
v² = u²-2gH
Given
u = 60m/s
v = 0m/s
g = ?
H = 72m
Substituting the given value into the equation;
0² = 60²-2g(72)
0 = 3600-144g
-3600 = -144g
g = -3600/-144
g = 25m/s²
The magnitude of his acceleration due to gravity on the planet is 25m/s²
<span>3598 seconds
The orbital period of a satellite is
u=GM
p = sqrt((4*pi/u)*a^3)
Where
p = period
u = standard gravitational parameter which is GM (gravitational constant multiplied by planet mass). This is a much better figure to use than GM because we know u to a higher level of precision than we know either G or M. After all, we can calculate it from observations of satellites. To illustrate the difference, we know GM for Mars to within 7 significant figures. However, we only know G to within 4 digits.
a = semi-major axis of orbit.
Since we haven't been given u, but instead have been given the much more inferior value of M, let's calculate u from the gravitational constant and M. So
u = 6.674x10^-11 m^3/(kg s^2) * 6.485x10^23 kg = 4.3281x10^13 m^3/s^2
The semi-major axis of the orbit is the altitude of the satellite plus the radius of the planet. So
150000 m + 3.396x10^6 m = 3.546x10^6 m
Substitute the known values into the equation for the period. So
p = sqrt((4 * pi / u) * a^3)
p = sqrt((4 * 3.14159 / 4.3281x10^13 m^3/s^2) * (3.546x10^6 m)^3)
p = sqrt((12.56636 / 4.3281x10^13 m^3/s^2) * 4.458782x10^19 m^3)
p = sqrt(2.9034357x10^-13 s^2/m^3 * 4.458782x10^19 m^3)
p = sqrt(1.2945785x10^7 s^2)
p = 3598.025212 s
Rounding to 4 significant figures, gives us 3598 seconds.</span>
