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”.



∴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: 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=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)

gravitation rocket

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


∴ 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)


  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).



  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



Join Discussion Forum and do your assignment: Find questions at the end of each lesson, Click here to discuss your answers in the forum

For advert placement/partnership, write

Download our free Android Mobile application: Save your data when you use our free app. Click picture to download. No subscription. stoplearn

We are interested in promoting FREE learning. Tell your friends about Click the share button below!