Critical velocity is the minimum velocity of projection for which satellite revolves in a circular orbit around the earth.

G = Universal Gravitational Constant

g_h = acceleration due to gravity at height h above earth’s surface

M= Mass of the earth

R =Radius of the earth

h = height at which satellite is orbiting

If satellite is revolving very close to the earth’s surface, then

So V_c= √ g_h(R+h) is the formula for Critical Velocity of Satellite Orbiting at height h from the Earth’s Surface

Let us consider mass of the earth as M, radius of earth as R, and the body is at an altitude h from the surface of the earth at distance r, gh is gravity at height h from the earth’s surface. The acceleration gained by an object because of gravitational force is called its acceleration due to gravity.

Acceleration due to gravity on the surface of the earth= g =  ………..(1)

Gravity at height h from the earth’s surface = gh=

Distance from surface of earth (r) = Radius of earth + height from the surface of he earth

r = R + h

Therefore,

By dividing (1) / (2),the formula for acceleration due to gravity at height h from the earth’s surface is given as,

G = Universal Gravitational Constant R = radius of the earth

M = mass of the earth

Density is defined as its mass divided by its volume. In the same way density of nucleus is defined as the ratio of mass of nucleus to its spherical volume. The mass of nucleus is equal to number of nucleons in nucleus (A) and the mass of single nucleon (m). Since mass of neutron and proton are same.

M = A m.

Nucleus is spherical in shape with radius r. So volume of nucleus (V) = 4/3(Π r³).
So Formula of density of a nucleus.

Where A = number of nucleons in nucleus

m = mass of single nucleon

r = Radius of nucleus.

Experiments that studies the size of nuclei indicate that the volume of nuclei is directly proportional to the total number of nucleons present.

V_n ∝ A  

where V_n = volume of nuclei and A = mass number

For sphere, V ∝ r³

where r = radius of a nucleus

Substituting value of V in above equation we get,

r³ ∝ A

So, r ∝ A^1/3

Using R₀ as a proportionality constant we get,

r = A^1/3 R₀

where A = mass number, r = radius of a nucleus and R₀ = proportionality constant.

R₀ = 1.25 fm (femtometre) with typical deviations of up to 0.2 fm from this value.

Atomic mass unit is taken to be the 1/12th of the actual mass of carbon isotope 6C12
The value of Atomic mass unit (a.m.u.) is given as,
1 a.m.u. = 1.66 × 10-27 kg
1 a.m.u. = 931 × 106 eV
1 a.m.u. = 931 MeV