Gravitational Force between Two Masses (Newton’s Law of Universal Gravitation)

Subject: 

PHYSICS

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Term:

FIRST TERM

Week:

WEEK 1

Class:

SS 3

Topic:

Gravitational Force between Two Masses (Newton’s Law of Universal Gravitation)

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Previous lesson: 

The pupils have previous knowledge of

 PHYSICS FIRST TERM EXAMINATION SS 3

that was taught as a topic in the previous lesson

Behavioural objectives:

At the end of the lesson, the learners will be able to

• State Newton’s law of gravitation.
• Define escape velocity.
• Write down the mathematical expression for:
  • Gravitational force.
  • Gravitational potential.

Instructional Materials:

• Wall charts
• Pictures
• Related Online Video
• Flash Cards

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Methods of Teaching:

• Class Discussion
• Group Discussion
• Asking Questions
• Explanation
• Role Modelling
• Role Delegation

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Reference Materials:

• Scheme of Work
• Online Information
• Textbooks
• Workbooks

Content:

Gravitational Force between Two Masses (Newton’s Law of Universal Gravitation)

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 F_g is the gravitational force in Newton,N,

Gis the universal or gravitational constant of value 6.7 × 10^-11 and expressed in Nm²/kg² 

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.

Gravitational Potential V_g

The gravitational potential V_g at 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, V_g decreases 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 V₀ as 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

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

2. Find the gravitational force between a proton and an electron if their distance of separation is  10^-10m. (m_p = 1.67 × 10^-27kg, m_e = 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

Presentation

The topic is presented step by step

Step 1:

The class teacher revises the previous topics

Step 2.

He introduces the new topic

Step 3:

The class teacher allows the pupils to give their own examples and he corrects them when the needs arise

Satellites and Parking Orbits

Satellites are bodies, natural or artificial which move in orbits around the moon or planets.Artificial  satellites are made by man.
Consider a satellite of mass m moving round the earth of mass M in an orbit as shown below:

If R is the radius of the earth, r is the radius of the orbit and v is the velocity with which the satellite is moving,

We have that centripetal force (due to the satellite) = gravitational force (due to the earth)

∴mv2r=GMmR2=mg∴mv2r=mg∴v2=gr

Hence, v=gr−−√ ——-(9)

To get the time for one complete revolution, i.e, period, we use

speed=distancetime

Therefore, the period T is given as:

T=length of circumferencespeed=2πRv

But v=gr−−√∴T=2πRgR√

By squaring both sides, we have,

T2=4π2R2gR√=4π2Rg

Hence, T=2πRg−−√ ——-(10)

Since R = 6400km and g = 9.8m/s², therefore, T = 5077.58s = 84.6mins. This is the period for an artificial satellite in the orbit above the earth.

If the period of the satellite in its orbit is equal to the period of the earth about its axis, which is 24hours, the satellite will stay at the same place above the earth as the earth rotates. Such orbit is called ‘’Parking Orbit’’.

Now, mv2R=GMmR2

But GM=gr2(fromGMr=gr=Vg)

We have, mv2R=mgrR2

Now, multiplying through by R/m, we have:

v2=gr2R

Again, v=2πRT

∴(2πRT)2=gr2R4π2R2T2=gr2R

By simplification,

R=gT2R24π2−−−−−√3 ——-(11) Radius of orbit

Hence, the length the satellite is above the earth surface l=Rorbit−Rearth

EVALUATION

1. State the relationship between radius of the orbit and the radius of the earth.
2. What is a satellite?
3. What is the value of g in orbit?

Energy of the Satellite

Supposing a satellite of mass m is revolving in an orbit of radius r round the earth of mass M with  a velocity v,then it will have both kinetic and potential energies.

Recall that mv2r=GMmR2=mg∴mv2r=GMmr2

Hence, v2=GMr

Now, k.e=12mV2

Putting in the value of v² above, we have:

k.e=12m×GMr=GMm2r

Hence, k.e=GMm2r ——-(12)

Also, p.e.=mgh=−GMmr

∴ Total energy k.e+p.e=GMm2r+−GMmr

Hence, total energy in orbit =−GMm2r ——-(13)

Worked Examples

1. Find the satellite orbit above the earth’s surface if the radius of that orbit is 6.5 × 10⁶m.

Solution:        

l=Rorbit−Rearthl=6.5×106−6.4×106=100,000m=100km

2. Determine the speed of revolution of a satellite round an orbit revolving at 85mins per revolution if the satellite is at 110km above the earth’s surface.

Solution:

Speed=2πRT

Now, T = 85mins = 5100s,

R = radius of the orbit =110km+6400km=6510=6,510,000mSpeed=2π×65100005100=8020.3m/s

Kepler’s Laws

Johannes Kepler in his working with data together with Tycho Brahe, put forth three laws which described the motion of the planets across the sky.

The laws are:

1. The Law of Orbits: All planets move in elliptical orbits, with the sun at one focus.
2. The Law of Areas: Aline that connects a planet to the sun sweeps out equal areas in equal
3. The Law of Periods: The square of the period of any planet is directly proportional to the cube of the semi-major axis of its orbit.

Kepler’s laws were derived for orbits around the sun, but they apply to satellite orbits as well.

GENERAL EVALUATION

1. State the mathematical expression of Isaac Newton’s law of gravitation.
2. What is the mathematical relationship between gravitational intensity and gravitational potential?
3. State the Kepler’s laws.
4. Considering a rocket launched from the earth’s surface, show that the velocity of escape V₀ is  given by:  V0=2gR−−−−√
5. Estimate the possible gravitational force between the earth and the moon at 4 × 10⁸m apart if  their masses are 6 ×10²⁴kg and 7 ×10²²kgrespectively.