Orbital Period Graphs | CAIE 9702 Paper 4 Notes

Chapters

Paper 4

Physics9702

Orbital Period Graphs

A graph question may use height above the surface instead of orbit radius from the centre.

Rewrite the period law into the straight-line form shown by the axes before reading the gradient and intercept.

Exam graph: T²ᐟ³ against height h

Height $h$ is measured from the surface. Orbit radius is measured from the centre. Here, $R$ is the planet radius.
$r=h+R$
At zero height, the orbit radius still equals the planet radius. Therefore, the graph has a positive vertical . It does not pass through the origin.
The horizontal axis is $h/10^6\ \text{m}$ . A reading of 12 must be changed back to metres before calculating the gradient.
$h=12\times10^6\ \text{m}$

The positive intercept and gradient are used to find the planet radius and mass.

Replace the orbit radius with height plus planet radius (9702/41/O/N/24 Q1(b)(ii)).
$(h+R)^3=\dfrac{GMT^2}{4\pi^2}$
Take the cube root and make $T^{2/3}$ the subject.
$T^{2/3}=\left(\dfrac{4\pi^2}{GM}\right)^{1/3}(h+R)$
Write the constant factor as k. This gives a straight-line equation.
$k=\left(\dfrac{4\pi^2}{GM}\right)^{1/3}$
$T^{2/3}=kh+kR$
The gradient is $k$ and the intercept is $kR$ .

Worked example

Use the gradient and intercept to find M and R

A line on a graph of T²ᐟ³ against h passes through (0, 360) and (12 × 10⁶ m, 1280). Find the planet mass M and radius R.

Calculate the gradient using height in metres.
$k=\dfrac{1280-360}{12\times10^6}$
$k=7.7\times10^{-5}\ \text{s}^{2/3}\text{m}^{-1}$
Use the gradient to find the planet mass.
$M=\dfrac{4\pi^2}{Gk^3}$
$M=1.3\times10^{24}\ \text{kg}$
Divide the intercept by the gradient to find the planet radius (9702/41/O/N/24 Q1(b)(iii)).
$R=\dfrac{360}{7.7\times10^{-5}}$
$R=4.7\times10^6\ \text{m}$

Answer

M=1.3\times10^{24}\ \text{kg}

Answer

R=4.7\times10^6\ \text{m}

Common mistake

Do not put height

h

into an equation that needs orbit radius

r

. Add the planet radius first. A positive intercept is expected.

On a T²ᐟ³ against h graph, the gradient gives the planet mass. Divide the intercept by the gradient to find the planet radius.

Your turnquick check

A graph of $T^{2/3}$ against height h has gradient k and vertical intercept b. Does it pass through the origin, and how do you find the planet radius?

Show worked answer

It does not pass through the origin because an orbit at h = 0 still has radius R. The intercept is:

b=kR

Therefore, the radius is:

R=\dfrac{b}{k}

Orbital Period Graphs

A graph question may use height above the surface instead of orbit radius from the centre.

Rewrite the period law into the straight-line form shown by the axes before reading the gradient and intercept.

Exam graph: T²ᐟ³ against height h

Height $h$ is measured from the surface. Orbit radius is measured from the centre. Here, $R$ is the planet radius.
$r=h+R$
At zero height, the orbit radius still equals the planet radius. Therefore, the graph has a positive vertical . It does not pass through the origin.
The horizontal axis is $h/10^6\ \text{m}$ . A reading of 12 must be changed back to metres before calculating the gradient.
$h=12\times10^6\ \text{m}$

The positive intercept and gradient are used to find the planet radius and mass.

Replace the orbit radius with height plus planet radius (9702/41/O/N/24 Q1(b)(ii)).
$(h+R)^3=\dfrac{GMT^2}{4\pi^2}$
Take the cube root and make $T^{2/3}$ the subject.
$T^{2/3}=\left(\dfrac{4\pi^2}{GM}\right)^{1/3}(h+R)$
Write the constant factor as k. This gives a straight-line equation.
$k=\left(\dfrac{4\pi^2}{GM}\right)^{1/3}$
$T^{2/3}=kh+kR$
The gradient is $k$ and the intercept is $kR$ .

Worked example

Use the gradient and intercept to find M and R

A line on a graph of T²ᐟ³ against h passes through (0, 360) and (12 × 10⁶ m, 1280). Find the planet mass M and radius R.

Calculate the gradient using height in metres.
$k=\dfrac{1280-360}{12\times10^6}$
$k=7.7\times10^{-5}\ \text{s}^{2/3}\text{m}^{-1}$
Use the gradient to find the planet mass.
$M=\dfrac{4\pi^2}{Gk^3}$
$M=1.3\times10^{24}\ \text{kg}$
Divide the intercept by the gradient to find the planet radius (9702/41/O/N/24 Q1(b)(iii)).
$R=\dfrac{360}{7.7\times10^{-5}}$
$R=4.7\times10^6\ \text{m}$

Answer

M=1.3\times10^{24}\ \text{kg}

Answer

R=4.7\times10^6\ \text{m}

Common mistake

Do not put height

h

into an equation that needs orbit radius

r

. Add the planet radius first. A positive intercept is expected.

On a T²ᐟ³ against h graph, the gradient gives the planet mass. Divide the intercept by the gradient to find the planet radius.

Your turnquick check

A graph of $T^{2/3}$ against height h has gradient k and vertical intercept b. Does it pass through the origin, and how do you find the planet radius?

Show worked answer

It does not pass through the origin because an orbit at h = 0 still has radius R. The intercept is:

b=kR

Therefore, the radius is:

R=\dfrac{b}{k}