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PHYSICS, M5 2023 HSC 14 MC

Planet \(X\) has a mass 4 times that of Earth and a radius 3 times that of Earth. The escape velocity at the surface of Earth is 11.2 km s\(^{-1}\).

What is the escape velocity at the surface of planet \(X\) ?

  1. 8.40 km s\(^{-1}\) 
  2. 9.70 km s\(^{-1}\)
  3. 12.9 km s\(^{-1}\)
  4. 14.9 km s\(^{-1}\)
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\(C\)

Show Worked Solution

→ The escape velocity of Earth, \( v_\text{esc}=  \sqrt{ \dfrac{2GM}{r}} \)

→ The escape velocity of planet \(X\)

\(v_\text{esc}\) \(=\sqrt{ \dfrac{2G \times 4M}{3r}} \)  
  \(=\dfrac{2}{\sqrt{3}} \times \sqrt{ \dfrac{2GM}{r}} \)  
  \(=\dfrac{2}{\sqrt{3}} \times 11.2 \)  
  \(= 12.9\ \text{km s}^{-1}\)  

 

\(\Rightarrow C\)

PHYSICS, M5 EQ-Bank 28

A bullet is fired vertically from the surface of Mars, at the escape velocity of Mars. Another bullet is fired vertically from the surface of Earth, at the escape velocity of Earth.

Neglecting air resistance, compare the energy transformations of the two bullets.  (5 marks)

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→ The total energy of the bullet fired from the surface of Mars will remain constant. Its kinetic energy will transform into potential energy as it gains altitude. 

→ As the bullet fired from Mars reaches an ‘infinite’ distance away, it escapes Mars’ gravitational attraction. Here, the kinetic energy of the bullet is zero as it has expended all of its initial kinetic energy.

→ The potential energy (`U`) of the bullet has also decreased to zero as  `U=-(GMm)/(r)`.

→ A similar process occurs for the bullet fired from the surface of Earth at its escape velocity. As the mass and radius of Earth are different to that of Mars, the actual escape velocity, `v_(esc)=sqrt((2GM)/(r))` will be different.

→ The initial values of kinetic and potential energy, `U=-(GMm)/(r)` will also be different.

Show Worked Solution

→ The total energy of the bullet fired from the surface of Mars will remain constant. Its kinetic energy will transform into potential energy as it gains altitude. 

→ As the bullet fired from Mars reaches an ‘infinite’ distance away, it escapes Mars’ gravitational attraction. Here, the kinetic energy of the bullet is zero as it has expended all of its initial kinetic energy.

→ The potential energy (`U`) of the bullet has also decreased to zero as  `U=-(GMm)/(r)`.

→ A similar process occurs for the bullet fired from the surface of Earth at its escape velocity. As the mass and radius of Earth are different to that of Mars, the actual escape velocity, `v_(esc)=sqrt((2GM)/(r))` will be different.

→ The initial values of kinetic and potential energy, `U=-(GMm)/(r)` will also be different.

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 2017 HSC 24

The escape velocity from a planet is given by  `v = sqrt((2GM)/(r))`.

  1. The radius of Mars is  `3.39 × 10^(6) \ text{m}`  and its mass is  `6.39 × 10^(23) \ text{kg}`.
  2. Calculate the escape velocity from the surface of Mars.   (2 marks)
  1. Using the law of conservation of energy, show that the escape velocity of an object is independent of its mass.   (3 marks)
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a.   `v=5010  text{m s}^(-1)`

b.   Applying the law of conservation of energy:

→ The object’s initial mechanical energy must equal its final mechanical energy.

`KE_(i)+U_(i)=KE_(f)+U_(f)`

→ The escape velocity is the minimum velocity required for an object to escape from a central mass and never return.

→ As an object reaches an infinite distance away, `U_(f)=0`

→ When an object has just enough speed to escape, its final speed is zero, hence, `KE_(f)=0`.

→ It follows

`KE_(i)+U_(i)` `=0`  
`(1)/(2)mv^(2)-(GMm)/(r)` `=0`  
`mv^(2)` `=(2GMm)/(r)`  
`∴ v_(esc)` `=sqrt((2GM)/(r))`  

 
→ Which is independent of the object’s mass.

Show Worked Solution
a.    `v` `=sqrt((2xx6.67 xx10^(-11)xx6.39 xx10^(23))/(3.39 xx10^(6)))`
    `=5014.5  text{m s}^(-1)`
    `=5015  text{m s}^(-1)\ \ text{(to 0 d.p.)}`

 

b.   Applying the law of conservation of energy:

→ The object’s initial mechanical energy must equal its final mechanical energy.

`KE_(i)+U_(i)=KE_(f)+U_(f)`

→ The escape velocity is the minimum velocity required for an object to escape from a central mass and never return.

→ As an object reaches an infinite distance away, `U_(f)=0`

→ When an object has just enough speed to escape, its final speed is zero, hence, `KE_(f)=0`.

→ It follows

`KE_(i)+U_(i)` `=0`  
`(1)/(2)mv^(2)-(GMm)/(r)` `=0`  
`mv^(2)` `=(2GMm)/(r)`  
`∴ v_(esc)` `=sqrt((2GM)/(r))`  

 
→ Which is independent of the object’s mass.


Mean mark (b) 53%.

PHYSICS, M5 2020 HSC 12 MC

In which of the following would the satellite have the greatest escape velocity from Earth?
 

 

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`D`

Show Worked Solution

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

→ escape velocity is independent of the satellite’s mass

`v_(esc) prop (1)/(sqrt(r))`

→ the smaller the radius, the higher the required escape velocity

`=>D`