Newton’s law of gravitation IB DP Physics Study Notes
Newton’s law of gravitation IB DP Physics Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with Students should understand
Kepler’s three laws of orbital motion
Newton’s universal law of gravitation as given by $F = G\frac{m_1m_2}{r^2}$ for bodies treated as point masses
conditions under which extended bodies can be treated as point masses
that gravitational field strength g at a point is the force per unit mass experienced by a small point
mass at that point as given by $g = \frac{F}{m}= G\frac{M}{r^2}$gravitational field lines.
Standard level and higher level: 2 hours
Additional higher level: 2 hours
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Kepler’s Laws – 1st and 2nd Law & 3rd Law
Kepler’s 1st Law
- Planets follow elliptical orbits with the star at one focus of the elipse.
Kepler’s 2nd Law
- A line joining the planet and star will sweep the same area during the same amount of time, regardless of position in the orbit.
Kepler’s Laws – 3rd Law
- Derive Kepler’s law, which states that the period T of an object in a circular orbit about a body of mass M is given by \(T^{2}=[\frac{4\pi^{2}}{GM}]r^{3}\).
SOLUTION:
In circular orbit \(F_{C}=ma_{c}\).
•From Newton’s law of gravitation \(F_{C}=\frac{GMm}{r^{2}}\).
•From Topic A.2, \(a_{C}=\frac{4\pi^{2}r}{T^{2}}\).
Then
$ma_{c}=\frac{GMm}{r^{2}}$
$\frac{4\pi^{2}r}{T^{2}}=\frac{GM}{r^{2}}$
$4\pi^{2}r^{3}=GMT^{2}$
\(T^{2}=[\frac{4\pi^{2}}{GM}]r^{3}\).
EXAMPLE:
A satellite in geosynchronous orbit takes 24 hours to orbit the earth. Thus, it can be above the same point of the earth’s surface at all times, if desired. Find the necessary orbital radius, and express it in terms of earth radii. \(R_E = 6.37 \times 10^6\) m.
▶️Answer/Explanation
SOLUTION:
\(T = (24h)(3600 s h^{-1}) = 86400\) s.
Then from Kepler’s law \(T^2 = [\frac{4\pi^2}{GM}] r^3\) we have
\(r^3 = \frac{T^2}{[\frac{4\pi^2}{GM}]} = \frac{86400^2}{[\frac{4\pi^2}{(6.67 \times 10^{-11})(5.98 \times 10^{24})}]} = 7.54 \times 10^{22}\)
\(r = (42250474 \text{ m}) (\frac{1 R_E}{6.37 \times 10^6 \text{ m}}) = 6.63 R_E\)
EXAMPLE:
A satellite in geosynchronous orbit takes 24 hours to orbit the earth. Thus, it can be above the same point of the earth’s surface at all times, if desired. Find the necessary orbital radius, and express it in terms of earth radii. \(R_E = 6.37 \times 10^6\) m.
▶️Answer/Explanation
SOLUTION:
\(T = (24h)(3600 s h^{-1}) = 86400\) s.
Then from Kepler’s law \(T^2 = [\frac{4\pi^2}{GM}] r^3\) we have
\(r^3 = \frac{T^2}{[\frac{4\pi^2}{GM}]} = \frac{86400^2}{[\frac{4\pi^2}{(6.67 \times 10^{-11})(5.98 \times 10^{24})}]} = 7.54 \times 10^{22}\)
\(r = (42250474 \text{ m}) (\frac{1 R_E}{6.37 \times 10^6 \text{ m}}) = 6.63 R_E\)
Newton’s law of gravitation
∙In 1687 Isaac Newton published what has been called by some the greatest scientific discovery of all time – his universal law of gravitation.
∙The law states that the gravitational force between two point masses m1 and m2 is proportional to their product, and inversely proportional to the square of their separation r.
∙The actual value of G, the universal gravitational constant, was not known until Henry Cavendish conducted a tricky experiment in 1798 to find it.
Be very clear that r is the distance between the centers of the masses.
FYI
∙The radius of each mass is immaterial.
Gravitational field strength
∙Suppose a mass m is located a distance r from a another mass M.
∙The gravitational field strength g is the force per unit mass acting on m due to the presence of M. Thus
The units are newtons per kilogram (N kg$^{ -1}$).
Note that from Newton’s second law, F = ma, we see that a N kg$^{ -1}$ is also a m s$^{ -2}$, the units for acceleration.
Note further that weight has the formula F = mg, and that the g in this formula is none other than the gravitational field strength!
On the earth’s surface, g = 9.8 N kg$^{ -1}$ = 9.8 m s$^{ -2}$.
- Suppose a mass m is located on the surface of a planet of radius R. We know that its weight is F = mg.
- But from the law of universal gravitation, the weight of m is equal to its attraction to the planet’s mass M and equals
- F = (GMm) / R².
- Thus
- mg = (GMm) / R².
- GM / R² = g gravitational field strength at the surface of a planet of mass M and radius R
- This same derivation works for any r.
-
Gravitational field lines.
• Compare the gravitational force formula F = (GMm)/r² (Force – action at a distance)
with the gravitational field formula g = (GM)/r² (Field – local curvature of space)
• Note that the force formula has two masses, and the force is the result of their interaction at a distance r.
• Note that the field formula has just one mass – namely the mass that “sets up” the local field in the space surrounding it. It “curves” it.
• The field view of the universe (spatial disruption by a single mass) is currently preferred over the force view (action at a distance) but we will not get into this topic.
∙Consider the gravitational field of the sun. If we consider the field lines to represent gravitational field strength, our sketch of the gravitational field is vastly simplified:
∙In fact, we don’t even have to draw the sun – the arrows are sufficient to denote its presence.
∙To simplify field drawings even more, we take the convention of drawing “field lines” as a single arrow.
∙In the first sketch the strength of the field at a point is determined by the length of the field arrows in the vicinity of that point.
∙The second sketch has single arrows, so how do we know how strong the field is at a particular point in the vicinity of a mass?
∙We simply look at the concentration of the field lines. The closer together the field lines, the stronger the field.
∙In the red region the field lines are closer together than in the green region.
∙Thus the red field is stronger than the green field.