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PHYSICS, M5 2022 VCE 2

There are over 400 geostationary satellites above Earth in circular orbits. The period of orbit is one day (86 400 seconds). Each geostationary satellite remains stationary in relation to a fixed point on the equator. Figure 2 shows an example of a geostationary satellite that is in orbit relative to a fixed point, \(\text{X}\), on the equator.
 

  1. Explain why geostationary satellites must be vertically above the equator to remain stationary relative to Earth's surface.  (2 marks)
  1.  Using  \(G=6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}, M_{\text{E}} = 5.98 \times 10^{24}\ \text{kg}\)  and  \(R_{\text{E}} = 6.37 \times 10^{6}\ \text{m}\), show that the altitude of a geostationary satellite must be equal to \(3.59 \times 10^{7}\ \text{m}\).  (4 marks)
  1. Calculate the speed of an orbiting geostationary satellite.  (3 marks)

Show Answers Only

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.

b.    See worked solutions

c.    3076 ms\(^{-1}\)

Show Worked Solution

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.
 

♦♦♦ Mean mark (a) 19%.
b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(r\) \(=\sqrt[3]{\dfrac{GMT^2}{4\pi^2}}\)
    \(=\sqrt[3]{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (86\ 400)^2}{4\pi^2}}\)
    \(=42.250 \times 10^6\ \text{m}\)

 
\(\therefore \ \text{Altitude}\ =42.250 \times 10^6 -6.37 \times 10^6 =3.59 \times 10^7\ \text{m}\)
 

c.     \(v\) \(=\sqrt{\dfrac{GM}{r}}\)
    \(=\sqrt{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42.25 \times 10^6}}\)
    \(=3073\ \text{ms}^{-1}\)

PHYSICS, M5 2023 HSC 5 MC

An exoplanet is in an elliptical orbit, moving in the direction shown. The distances between consecutive positions \(P, Q, R\) and \(S\) are equal.
 


 

Between which two points is the exoplanet's travel time the greatest?

  1. \(P\) and \(Q\)
  2. \(Q\) and \(R\)
  3. \(R\) and \(S\)
  4. \(S\) and \(P\)
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\(D\)

Show Worked Solution

→ As the exoplanet is in an elliptical orbit it will travel the slowest when it is the furthest away from the star.

→ As all distances between the points are equal, the exoplanet will have the greatest travel time between \(S\) and \(P\) where it moves the slowest.

\(\Rightarrow D\)

PHYSICS, M5 EQ-Bank 29

A student used the following scale diagram to investigate orbital properties. The diagram shows a planet and two of its moons, `V` and `W`. The distances between each of the moons and the planet are to scale while the sizes of the objects are not.
 

Complete the table to compare the orbital properties of Moon `V` and Moon `W`. Show relevant calculations in the space below the table.  (4 marks)
 

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\begin{array} {|l|c|c|c|}
\hline   & \textit{Orbital radius} & \textit{Orbital period} & \textit{Orbital velocity} \\
\hline \text{Quantitative Comp.} & 3.0 & 5.2 & 0.58 \\
\hline \text{Qualitative Comp.} & \text{Larger} & \text{Larger} & \text{Slower} \\
\hline \end{array}

 
→ Comparing radii:

`(r_(w))/(r_(v))=(10)/(3.3)=3`
 

→ As moons `V` and `W` are both orbiting the same body, the ratio `(r^3)/(T^2)`will be the same for both.

→ Comparing orbital periods:

`(r_(w)^(3))/(T_(w)^(2))` `=(r_(v)^(3))/(T_(v)^(2))`  
`((T_(w))/(T_(v)))^(2)` `=((r_(w))/(r_(v)))^(3)=3^3`  
`(T_(w))/(T_(v))` `=sqrt(27)=5.2`  

 
→ Comparing orbital velocities.

`v=(2pir)/(T)`

 
→ Using the calculations for radius and period:

`v_(w)` `=(2pir_(w))/(T_(w))`  
  `=(2pi(3.0xxr_(v)))/(5.2xxT_(v))`  
  `=(3.0)/(5.2)xx(2pir_(v))/(T_(v))`  
  `=(3.0)/(5.2)xxv_(v)`  
`(v_(w))/(v_(v))` `=0.58`  
Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline   & \textit{Orbital radius} & \textit{Orbital period} & \textit{Orbital velocity} \\
\hline \text{Quantitative Comp.} & 3.0 & 5.2 & 0.58 \\
\hline \text{Qualitative Comp.} & \text{Larger} & \text{Larger} & \text{Slower} \\
\hline \end{array}

 
→ Comparing radii:

`(r_(w))/(r_(v))=(10)/(3.3)=3`
 

→ As moons `V` and `W` are both orbiting the same body, the ratio `(r^3)/(T^2)`will be the same for both.

→ Comparing orbital periods:

`(r_(w)^(3))/(T_(w)^(2))` `=(r_(v)^(3))/(T_(v)^(2))`  
`((T_(w))/(T_(v)))^(2)` `=((r_(w))/(r_(v)))^(3)=3^3`  
`(T_(w))/(T_(v))` `=sqrt(27)=5.2`  

 
→ Comparing orbital velocities.

`v=(2pir)/(T)`

 
→ Using the calculations for radius and period:

`v_(w)` `=(2pir_(w))/(T_(w))`  
  `=(2pi(3.0xxr_(v)))/(5.2xxT_(v))`  
  `=(3.0)/(5.2)xx(2pir_(v))/(T_(v))`  
  `=(3.0)/(5.2)xxv_(v)`  
`(v_(w))/(v_(v))` `=0.58`  

PHYSICS, M5 EQ-Bank 17 MC

Two identical masses are placed at points `P` and `Q`. The escape velocity and circular orbital velocity of the mass at point `P` are `v_{P_(esc)}` and `v_{P_{o rb}}`. The escape velocity and circular orbital velocity of the mass at point `Q` are `v_{Q_(e s c)}` and `v_{Q_{o rb}}`. The diagram is drawn to scale and `X` denotes the centre of Earth.
 

The velocity for a body in circular orbit is given by  `v_{o rb} = sqrt((GM)/r`.

What is the value of `(v_{Q_(e s c)})/v_{P_{o rb}}`?

  1. `0.5`
  2. `1`
  3. `sqrt(2)`
  4. `2`
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`B`

Show Worked Solution

→ The escape velocity of an object is given by `v_(esc)=sqrt((2GM)/(r))`

→ As the diagram is to scale, it can be measured that the distance from `Q` to `X` is twice that from `P` to `X`.

→ Let  `r` = distance from `P` to `X`  and  `2r` = distance from `Q` to `X`:

→ `v_(Q_(esc))=sqrt((2GM)/(2r))=sqrt((GM)/(r))`

→ `v_(p_(text{orb}))=sqrt((GM)/(r))` 

→ Hence, `v_(Q_(esc))=v_(P_(text{orb}))` → `(v_(Q_(esc)))/(v_(p_(text{orb})))=1`

`=>B`

PHYSICS, M5 2015 HSC 26

Consider the following two models used to calculate the work done when a 300 kg satellite is taken from Earth's surface to an altitude of 200 km.

You may assume that the calculations are correct.
 

  1. What assumptions are made about Earth's gravitational field in models `X` and `Y` that lead to the different results shown?   (2 marks)
  1. Why do models `X` and `Y` produce results that, although different, are close in value?   (1 mark)
  1. Calculate the orbital velocity of the satellite in a circular orbit at the altitude of 200 km.   (3 marks)
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a.  Model `X`:

→ Assumes Earth’s gravitational field strength remains constant moving upwards from the surface.

Model `Y`:

→ Assumes Earth’s gravitational field strength changes with altitude.

b.   Similarity of results due to:

→ The variation in gravitational field strength from Earth’s surface to an altitude of 200 km is minimal, so both models `X` and `Y` produce similar results.

c.   `v=7797  text{ms}^(-1)`

Show Worked Solution

a.   Model `X`:

→ Assumes Earth’s gravitational field strength remains constant moving upwards from the surface.

Model `Y`:

→ Assumes Earth’s gravitational field strength changes with altitude.


♦ Mean mark (a) 54%.

b.   Similarity of results due to:

→ The variation in gravitational field strength from Earth’s surface to an altitude of 200 km is minimal, so both models `X` and `Y` produce similar results.


♦♦ Mean mark (b) 38%.

c.    Centripetal force = force due to gravity:

`F_(c)` `=F_(g)`  
`(mv^2)/(r)` `=(GMm)/(r^2)`  
`:.v` `=sqrt((GM)/(r))`  
  `=sqrt((6.67 xx10^(-11)xx6xx10^(24))/(6.58 xx10^(6)))`  
  `=7797  text{ms}^(-1)`  

PHYSICS M5 2022 HSC 13 MC

Two satellites share an orbit around a planet. One satellite has twice the mass of the other.
  


  

Which quantity would be different for the two satellites?

  1. Speed
  2. Momentum
  3. Orbital period
  4. Centripetal acceleration
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`B`

Show Worked Solution

The orbital velocity, `v=sqrt((GM)/(r))` is independent of the mass of the orbiting body.

→ Both satellites have the same orbital velocity.

→ The momentum,  `p=mv` of the satellite with mass 2`m` is twice that of the satellite with mass `m`.

→ The two satellites have different momentum.

`=>B`

PHYSICS M5 2022 HSC 10 MC

The orbital velocity, `v`, of a satellite around a planet is given by  `v=sqrt((GM)/(r))`.

Which graph is consistent with this relationship?
 

  

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`D`

Show Worked Solution

Squaring both sides of the given equation gives:

`v^2=(GM)/(r)`

→   `v^2 prop (1)/(r)`

→ A line of `v^2` plotted against `(1)/(r)` will yield a linear relationship.

`=>D`

PHYSICS, M5 2019 HSC 14 MC

A satellite in circular orbit at a distance `r` from the centre of Earth has an orbital velocity `v`.

If the distance was increased to `2r`, what would be the satellite's orbital velocity?

  1. `(v)/2`
  2. `0.7v`
  3. `1.4v`
  4. `2v`
Show Answers Only

`B`

Show Worked Solution

The satellites centripetal force is provided by gravity:

`F_(c)` `=F_(g)`  
`(mv^2)/(r)` `=(GMm)/(r^2)`  
`v` `=sqrt((GM)/(r))`  
`∴ v` `prop (1)/(sqrt(r))`  

 
Hence, doubling the satellites radius changes its velocity by a factor of `(1)/(sqrt(2))`

`=>B`


♦ Mean mark 42%.