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Motion in a gravitational field IB DP Physics Study Notes

Motion in a gravitational field IB DP Physics Study Notes - 2025 Syllabus

Motion in a gravitational field IB DP Physics Study Notes

Motion in a gravitational field 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

  • the gravitational field strength g as the gravitational potential gradient as given by \(g = -\frac{\Delta V_g}{\Delta r}\)

  • the work done in moving a mass m in a gravitational field as given by \(W = m\Delta V_g\)

  • equipotential surfaces for gravitational fields

  • the relationship between equipotential surfaces and gravitational field lines

  • the escape speed \(v_{esc}\) at any point in a gravitational field as given by \(v_{esc} = \sqrt{\frac{2GM}{r}}\)

  • the orbital speed \(v_{orbital}\) of a body orbiting a large mass as given by \(v_{orbital} = \sqrt{\frac{GM}{r}}\)

  • the qualitative effect of a small viscous drag force due to the atmosphere on the height and speed of an orbiting body.

Standard level and higher level: 8 hours
Additional higher level: 6 hours

IB DP Physics 2025 -Study Notes -All Topics

Potential gradient – gravitational

The gravitational potential gradient (GPG) is the change in gravitational potential per unit distance. Thus the GPG $=\frac{\Delta V_g}{\Delta r}$.

EXAMPLE:

Find the GPG in moving from Earth’s surface to 5 radii from Earth’s center.

▶️Answer/Explanation

SOLUTION: In a previous slide we showed that

$$
\begin{aligned}
& \Delta V_g=+5.01 \times 10^7 \mathrm{Jkg}^{-1} . \\
& \bullet r_1=6.37 \times 10^6 \mathrm{~m} . \\
& \bullet r_2=5\left(6.37 \times 10^6\right)=3.19 \times 10^7 \mathrm{~m} . \\
& \bullet \Delta r=r_2-r_1=3.19 \times 10^7-6.37 \times 10^6=2.55 \times 10^7 \mathrm{~m} . \\
& \quad \text { GPG }=\frac{V_g}{\Delta r}=\frac{5.01 \times 10^7}{2.55 \times 10^7}=1.96 \mathrm{~J} \mathrm{~kg}^{-1} \mathrm{~m}^{-1} .
\end{aligned}
$$

Potential gradient – gravitational

Earlier, we found that near Earth, $g=-\frac{\Delta v_g}{\Delta y}$.
The following potential gradient (which we will not prove) works at the planetary scale:

EXAMPLE:

The gravitational potential in the vicinity of a planet changes from ${ }^{-6.16 \times 10^7 \mathrm{~J} \mathrm{~kg}^{-1} \text { to }-6.12 \times 10^7 \mathrm{~J} \mathrm{~kg}^{-1} .}$ in moving from $1.80 \times 10^8 \mathrm{~m}$ to $2.85 \times 10^8 \mathrm{~m}$. What is the gravitational field strength in that region?

▶️Answer/Explanation

SOLUTION:

$$
\begin{aligned}
g=-\frac{\Delta V_g}{\Delta r} & =-\frac{-6.12 \times 10^7-\left(-6.16 \times 10^7\right)}{2.85 \times 10^8-1.80 \times 10^8} \\
g & =-\frac{4000000}{105000000}=-0.00381 \mathrm{~m} \mathrm{~s}^{-2}
\end{aligned}
$$

Work Done in Moving a Mass in a Gravitational Field

$
W = m \Delta V_g
$
where:
 \( W \) = Work done (Joules)
 \( m \) = Mass of the object (kg)
 \( \Delta V_g \) = Change in gravitational potential (J/kg)

The gravitational potential energy at a height \( h \) is given by:
$
U = mgh
$
where \( g \) is the acceleration due to gravity.

The gravitational potential at a point is defined as:

$
V_g = \frac{U}{m} = gh
$

When moving a mass \( m \) from one point to another in a gravitational field, the work done is given by the change in gravitational potential energy:
$
W = U_{\text{final}} – U_{\text{initial}}
$
Substituting \( U = mV_g \):
$
W = mV_{g,\text{final}} – mV_{g,\text{initial}}
$

$
W = m \Delta V_g
$
where \( \Delta V_g = V_{g,\text{final}} – V_{g,\text{initial}} \).

FYI

This formula shows that the work done in moving a mass in a gravitational field depends only on the mass and the change in gravitational potential, not the path taken.

Equipotential surfaces revisited – gravitational

Recall that equipotential surfaces  are imaginary surfaces at which the potential is the same.

Since the gravitational potential for a point mass is given by \(V_g = -\frac{GM}{r}\) it is clear that the equipotential surfaces are at fixed radii and hence are concentric spheres:

FYI

Generally equipotential surfaces are drawn so that the \(\Delta V_g\)s for consecutive surfaces are equal.

Because \(V_g\) is inversely proportional to r, the consecutive rings get farther apart as we get farther from the mass.

EXAMPLE:

Sketch the gravitational field lines around two point masses.

▶️Answer/Explanation

SOLUTION:

Remember that the gravitational field lines point inward, and that they are perpendicular to the equipotential surfaces.

 

Escape speed

∙We define the escape speed to be the minimum speed an object needs to escape a planet’s gravitational pull.
∙We can further define escape speed vesc to be that minimum speed which will carry an object to infinity and bring it to rest there.
∙Thus we see as r → ∞ then v → 0.

Note that escape speed is independent of the mass that is actually escaping!

From the conservation of mechanical energy we have $\Delta E_K+\Delta E_P=0$. Then

$$
\begin{aligned}
E_K-E_{K 0}+E_P-E_{P 0} & =0 \\
\left(\frac{1}{2}\right) m{v^2}-\left(\frac{1}{2}\right) m u^2+\left(-\frac{G M m}{r}\right)-\left(-\frac{G M m}{r_0}\right) & =0 \\
\left(\frac{1}{2}\right) m v_{\text {esc }}^2 & =\frac{G M m}{R}
\end{aligned}
$$

PRACTICE:

Find the escape speed from Earth.

▶️Answer/Explanation

SOLUTION:

$$
\begin{aligned}
& -M=5.98 \times 10^{24} \mathrm{~kg} \text { and } R=6.37 \times 10^6 \mathrm{~m} . \\
& v_{\text {esc }}^2=\frac{2 G M}{R}=\frac{2\left(6.67 \times 10^{-11}\right)\left(5.98 \times 10^{24}\right)}{6.37 \times 10^6} \\
& v_{\text {esc }}=11200 \mathrm{~ms}^{-1}(=40300 \mathrm{~km} / \mathrm{h}!)
\end{aligned}
$$

Orbital speed 

 An orbiting satellite has both kinetic energy and potential energy.

 The gravitational potential energy of an object of mass m in the gravitational field of Earth is \(E_p = -\frac{GMm}{r}\), where M is the mass of the earth.

 As we learned in Topic A, the kinetic energy of an object of mass m moving at speed v is \(E_K = \left(\frac{1}{2}\right)mv^2\).

 Thus the total mechanical energy of an orbiting satellite of mass m is 

EXAMPLE:

Show that the speed of an orbiting satellite having mass $m$ at a distance $r$ from the center of Earth $\left(\right.$ mass $M$ ) is $V_{\text {orbit }}=\sqrt{\frac{G M}{r}}$

▶️Answer/Explanation

SOLUTION:

In circular orbit $F_c=m a_c$ and $F_c=F_g=\frac{G M m}{r^2}$.
But $a_c=\frac{v^2}{r}$. Then

$$
\begin{aligned}
m a_c & =\frac{G M m}{r^2} \\
\frac{m v^2}{r} & =\frac{G M m}{r^2} \\
v^2 & =\frac{G M}{r} \\
v & =\sqrt{\frac{G M}{r}}
\end{aligned}
$$

EXAMPLE:

Show that the kinetic energy of an orbiting satellite having mass $m$ at a distance $r$ from the center of Earth (mass $M$ ) is $E_k=\frac{G M m}{2 r}$.

▶️Answer/Explanation

SOLUTION:

In circular orbit $F_c=m a_c$ and $F_c=\frac{G M m}{r^2}$

But $a_c=\frac{v^2}{r}$. Then

$$
\begin{aligned}
& m a_c=\frac{G M m}{r^2} \\
& \frac{m v^2}{r}=\frac{G M m}{r^2} \\
& m v^2=\frac{G M m}{r} \\
& \frac{1}{2} m v^2=\frac{G M m}{2 r}
\end{aligned}
$$

  

EXAMPLE:

Show that the total energy of an orbiting satellite at a distance $r$ from the center of Earth is

$$
E_T=-\frac{G M m}{2 r}
$$

▶️Answer/Explanation

SOLUTION: From $E=E_K+E_{\mathrm{P}}$ and the expressions for $E_{\mathrm{K}}$ and $E_{\mathrm{P}}$ we have

$$
\begin{aligned}
& E_T=E_K+E_P \\
& E_T=\frac{G M m}{2 r}-\frac{G M m}{r} \\
& E_T=\frac{G M m}{2 r}-\frac{2 G M m}{2 r} \\
& E_T=-\frac{G M m}{2 r}
\end{aligned}
$$

Orbital Drag

  • The atmosphere becomes thinner and thinner as you move away from the surface, but there are still atmospheric particles in low Earth orbit.

  • What effect will this have on the speed of objects in orbit? As the object hits the gas particles, it will transfer momentum to them, slowing down.

  • What effect will this have on the distance of the object from the surface of Earth? The gravitational force will pull the slower object toward the surface. (the object will start to fall!)

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