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# Gravitational Force between Two Masses (Newton’s Law of Universal Gravitation) and Escape Velocity

A gravitational field is a region of space surrounding a body that has the property of mass. Sir Isaac Newton, in 1666, propounded the universal law of gravitation.

The law states that, ” the force of attraction between two given particles of masses M and m is directly proportional to the product of the masses and inversely proportional to the square of their distance of separation”.

Mathematically,

Fg∝MmandFg∝1r2∴Fg∝Mmr2

∴Fg=GMmr2 ——-(1)

Hence, G=Fgr2Mm

Where Fis the gravitational force in Newton,N,

Gis the universal or gravitational constant of value 6.7 × 10-11 and expressed in Nm2/kg

The gravitational field strength, also called acceleration due to gravity, ‘g’ is given by:

g=Fgm ——-(2)

But Fg=GMmr2∴g=GMmr2÷m=GMmr2×1m

Hence, g=GMr2 ——-(3)

This is the relationship between gravitational constant G and acceleration due to gravity g.

Therefore, we define gravitational field strength ‘g’ as force per unit mass. It is a vector quantity.

# Newton Biography and Gravitation SLIDE SHOW:

https://www.slideshare.net/slideshow/embed_code/key/83wSoKHAIOBG5TNewton and his universal theory of gravitation from pacotomasvalientejorda

Gravitational Potential Vg

The gravitational potential Vat a point is the work done in taking a unit mass from infinity to that  point on the surface of the earth.

Vg=gr ——-(4)

Vg=GMr2×r

∴ Vg=GMr ——-(5)

Where M is the mass of the earth and r is the radius of the earth of value 6.4 × 106m or 6400km

At any point, distance r from the centre of the earth, the gravitational potential experienced by a body of mass m is given by:

∴Vg=−Gmr ——-(6)

Since the potential at infinity is taken to be zero. (i.e, ∆potential =0−Vg=−Vg=−Gmr

The negative sign indicates that potential at infinity is higher than the potential close to the mass, that is, Vdecreases as r increases.

Escape Velocity V0

Consider a rocket of mass m placed at the centre of the earth’s surface O. If it is fired from that  point so that it just escapes the earth’s gravitational field, it has a kinetic energy, k.e given as:

k.e=12mV20 ——-(7)

But work is done(WD) in taking this rocket to a distance R so great that the gravitational field is  negligibly weak.

∴ Work done =mg×distance

This work done must be equal to the kinetic energy of the rocket at the point of take off

Work done =12mV20

∴ Work done =mg×R=mgR

but gravitational intensity g=GMR2

Work done =m(GMR2)R

Work done =GMmR

But k.e = WD

∴12mV20=GMmR∴V20=2GMR

∴ V0=2GMR−−−−√ ——-(8)

But gR=GMR

Hence, V0=2gR−−−−√ ——-(8)

We thus define the escape velocity Vas the velocity which is sufficient enough for a body to just  escape the earth’s gravitational field.

Since g = 9.8m/s2, and R = 6,400,000m

Thus V0=2×9.8×6.4×106−−−−−−−−−−−−−−−√=11,200m/s(40,320km/h)

EVALUATION (POST YOUR ANSWERS USING THE QUESTION BOX BELOW FOR EVALUATION AND DISCUSSION):

1. State Newton’s law of gravitation.
2. Define escape velocity.
3. Write down the mathematical expression for :
4. Gravitational force. ii. Gravitational potential.

Worked Examples

1. Calculate the gravitation force of attraction between two planets of masses 1024kg and 1027kg separated by a distance of 1020(G = 6.67 × 10-11).

Solution:

F=Gm1m2r2=6.67×10−11×1024×1027(1020)2=6.67N

1. Find the gravitational force between a proton and an electron if their distance of separation is  10-10m. (m= 1.67 × 10-27kg, m= 9.11 × 10-31kg

Solution:

F=Gm1m2r2=6.67×10−11×1.67×10−27×9.11×10−31(10−10)2=1.01×10−47N