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# The gravitational force

Before we go in gravitational force as it was defined by Newton law, let first all say something about gravitational field .The gravitational field can simple be described as field force. This force attracts all objects near to the Earth’s surface. That is the Earth exerts attractive force on every object near it.

# Newton law of gravitational force

It state that the force of attraction between two bodies of mass M1, and M2 is directly proportional to the products of the two mass and inversely proportional to the square of distance separating them.

# $f=\frac { Gm1m2 }{ { r }^{ 2 } }$ $f=\frac { Gm1m2 }{ { r }^{ 2 } }$

Where G is gravitational universal constant, F = gravitational force on each particle m1, m2

It is the gravitational attraction that holds the moon in its orbit about the Earth, and also Earth and other planets in their orbits around the sun

The relationship between the acceleration due to gravity and gravitational constant G

Using Newton law which state that the force of gravity on a body of mass m is by $F=mg\quad thus,\quad \frac { GMm }{ { R }^{ 2 } } =\quad mg$ $therefore,\quad g\quad =\quad \frac { GM }{ { R }^{ 2 } }$

# Variation of acceleration due to gravitational force

Since acceleration due to gravity g is given as g = GM/R2, the acceleration due gravity is not the same everywhere, because , though gravitational constant G and the mass of the Earth M are constant, but R which is the distance  between the center of Earth and any point on the Earth surface varies along the Earth’s surface. For this reason, the value of g varies from place to place along the earth surface and also this is why the value of g is greater at pole than at equator because the diameter of equator is greater than pole diameter. The value of g is big where R is small and small where R is big.

# Variation in gravitational acceleration g with the altitude

The acceleration due to gravity g at sea level is give as $g\quad =\quad \frac { GM }{ { R }^{ 2 } } \quad \quad but\quad the\quad acceleration\quad due\quad to\quad gravity\quad at\quad heighth\quad is\quad give\quad as$ ${ g }^{ 1 }=\frac { GM }{ { (R\quad +\quad h) }^{ 2 } }$ $\frac { { g }^{ 1 } }{ g } \quad =\frac { { R }^{ 2 } }{ { (R\quad +\quad h) }^{ 2 } } \quad by\quad division$

dividing further both the numerator and denominator with R2 we have $\frac { { g }^{ 1 } }{ g } \quad =\frac { 1 }{ { (1\quad +\frac { h }{ R } ) }^{ 2 } } \quad$

If h is small compared to R, we we can write ${ g }^{ 1 }\quad =g(1\quad +{ (\frac { h }{ R } })^{ -2 }\quad =\quad g(1\quad +\frac { 2h }{ R }$

Thus the acceleration due gravity decrease with increase in altitude. And this is why a body experience weightlessness at certain height above the surface of Earth (altitude). Weightlessness is also experience when a body is in a lift or an aeroplane descending.

# Gravitational force Potential

It is defined as  work done in taking a unit mass from infinite to a point and is given as $V\quad =\quad -\frac { GM }{ r }$

where V is gravitational potential.  This is true on the surface of Earth, but if a body of mass, m, is moved from infinite to the Earth surface of R, the work done on the body is a gravitational potential, which is given as $Workdone\quad \quad =\quad \frac { GMm }{ r }$

# Velocity of escape

For a rocket that is fired to escape the influence of gravitational force, you must consider the work done which is the product of the mass, m of the Rocket and the gravitational potential  between finite and Earth surface. Which is give as $Workdone\quad \quad =\quad m\quad \times \frac { GM }{ R }$

If the rocket moves with velocity v, then the kinetic energy is give as $\frac { 1 }{ 2 } m{ v }^{ 2 }\quad =\quad work\quad done$ $\frac { 1 }{ 2 } m{ v }^{ 2 }\quad =\quad \quad m\quad \times \frac { GM }{ r } \quad therefore\quad ,\quad v\quad =\quad \sqrt { (\frac { 2GM }{ R } ) } =\quad velocity\quad of\quad escape$ $But\quad \frac { GM }{ { R }^{ 2 } } \quad =\quad g,\quad therefore\quad v=\quad \sqrt { (2gR) }$ 