Here are the answers to the exam questions: Q1. By using equations for potential and kinetic energy, derive the equation for escape velocity: Total energy at infinity (Etot) = Kinetic energy (Ek) 1/2mv^2 = GMm/r For an object to escape, Etot must be positive or zero. 1/2mv^2 = -GMm/r mv^2 = 2GM/r v = √(2GM/r) Q2. Calculate the escape velocity for the following planets: a) Mars: mass = 6.46 × 1023 kg, radius = 3.39 × 106 m Escape velocity

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Orbits & escape velocity 13/02/2024
Q1. Calculate (a) the orbital speed and
(b) period of the International Space Station.
Earth radius, R = 6400 km
Orbital height of the ISS, H = 300 km
Earth mass, M = 6.0 x 10 24 kg
G = 6.672 x 10 -11 N m 2 kg – 2
Q2. Calculate the orbital radius of an Earth
satellite having an orbital period of 24 hours.

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Q1 review
(a) Orbital speed
v = √ (GM / r)
BUT: r = R + H = (6400 + 300)km = 6700 km = 6.7 x 106 m
and so: v = √ (6.672 x 10 -11 x 6.0 x 10 24 / 6.7 x 106)
= √ (4.003 x 10 14 / 6.7 x 106)
= √ (5.975 x 10 7)
orbital speed = 7.73 x 103 ms-1 = 7.73 kms-1
(b) Orbital period
T = √ (4π2 r 3 / GM)
= √ (4π2 x (6.7 x 106)3 ) / (6.672 x 10 -11 x 6.0 x 10 24)
= √ (1.187 x 10 22 / 4.003 x 10 14)
= √ (2.966 x 10 7)
orbital period = 5.45 x 103 s = 90.8 minutes = 1h 30 mins

3. SciDoc
Q2 review
Earth mass, M = 6.0 x 10 24 kg; G = 6.672 x 10 -11 N m 2 kg - 2
T = √ (4π2 r 3 / GM)
rearranged becomes: r 3 = T 2 GM / 4π2
r 3 = (24 x 60 x 60)2 x (6.672 x 10 -11 x 6.0 x 10 24 ) / 4π2
r 3 = (86400)2 x (4.003 x 10 14) / 4π2
r 3 = (7.465 x 10 9 ) x (4.003 x 10 14) / 4π2
r 3 = 7.570 x 10 22
r = 4.23 x 10 7 m = 42 300 km
This is 35 900 km above the Earth’s surface

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• An artificial satellite is an object that
people have made and launched into
orbit using rockets.
• Satellites vary in size. The largest
artificial satellite is the International
Space Station (ISS). The main part of
this is as big as a large five-bedroom
house, but including solar panels, it is
as large as a rugby field.
What is an artificial satellite?

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• Altitudes of satellites above the Earth’s surface:
• Low Earth orbit (LEO) – from 200 to 2000 km, for example, the ISS orbits at 400
km with a speed of 28 000 km/h, and time for one orbit is about 90 minutes.
• Medium Earth orbit (MEO) – most MEO satellites are at an altitude of 20 000
km, and time for one orbit is 12 hours.
• Geostationary orbit (GEO) – 36 000 km above the Earth. Time for one orbit is
24 hours. This is to match the rotation of the Earth so that the satellite appears
to stay above the same point above the Earth’s surface. This is used for many
communications and weather satellites.
Types of orbit

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T/N: Types of orbit
LEO: 200-2000 km. ISS space station,
telecommunications.
MEO: 20,000 km. Used for GPS.
GEO: Communications/TV. Over equator.
Calculate the time period of
these satellites above the
surface of Earth (remember
radius of Earth is 6400 km!)
a) LEO 200 km above
Earth’s surface
b) MEO 20,000 km above
Earth’s surface
c) GEO 36,000 km above
Earth’s surface.
What happens to the period
as the radius of orbit
increase? Why?

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The escape velocity from a planet is the minimum velocity an object must be
given to escape from the planet when projected vertically from the surface.
The gravitational potential V at a distance r from the centre of the planet of
mass M is given by
𝑉 = −
𝐺𝑀
𝑟
To move an object of mass m from the surface to infinity, the work that must be
done is ∆𝑊 = 𝑚∆𝑉 =
𝐺𝑀𝑚
𝑟
Escape velocity
Stretch: How can we use this expression to obtain the velocity needed to
escape the Earth’s gravitational field (i.e. the escape velocity)?

8. SciDoc
For the object to escape, the kinetic energy must be
more than the work required:
1
2
𝑚𝑣2 ≥ ∆𝑊
Therefore
1
2
𝑚𝑣2 ≥
𝐺𝑀𝑚
𝑅
Escape velocity is then given by 𝑣esc =
2𝐺𝑀
𝑅
And by using 𝑔 =
𝐺𝑀
𝑅
, we obtain a final expression for
the escape velocity which is:
𝑣esc = 2𝑔𝑅
Escape velocity
Stretch: What is the
escape velocity for Earth?
(Earth’s radius = 6400 km)

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Consider a satellite of mass m in a circular orbit of radius r, about a spherical
planet or star of mass M. The speed of the satellite v is given by:
𝑣2 =
𝐺𝑀
𝑟
as derived last lesson.
What is the kinetic energy Ek of the satellite?
1
2
𝑚𝑣2 =
1
2
𝑚 ×
𝐺𝑀
𝑟
=
𝐺𝑀𝑚
2𝑟
Energy of orbiting satellite

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The satellite is at distance r from the centre of the planet. At this distance
The gravitational potential 𝑉 = −
𝐺𝑀
𝑟
.
The satellite’s GPE:
𝐸𝑝 = 𝑚𝑉 = −
𝐺𝑀𝑚
𝑟
Therefore the total energy of the satellite 𝐸tot = 𝐸𝑝 + 𝐸𝑘:
𝐸tot = −
𝐺𝑀𝑚
𝑟
+
𝐺𝑀𝑚
2𝑟
= −
𝐺𝑀𝑚
2𝑟
Energy of orbiting satellite
Stretch: Sketch a
graph of kinetic,
potential & total
energy of a
potential against
the radius of orbit.

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Therefore, for a satellite in a circular orbit of radius r, its total energy is equal to:
𝐸tot = −
𝐺𝑀𝑚
2𝑟
Energy of orbiting satellite

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Area under graphs
Work done = Force × Distance
The graph shows how the force of
gravity on a 1kg mass varies with
distance from infinity to the surface of a
planet. What does the area under the
curve represent?
The area under the curve represents the
work done to move the 1kg mass from
the surface to infinity.

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Area under graphs
What does the area under this graph represent? (look
at your equation sheet if it helps).
∆V from area under graph of g against r.

14. SciDoc
Gravitational fields recap
Gravitational field
strength
𝑔 =
𝐹
𝑚
= −
∆𝑉
∆𝑟
=
𝐺𝑀
𝑟2
Satellite motions 𝑟3
𝑇2
=
𝐺𝑀
4π2
Potential gradient
=
∆𝑉
∆𝑟
Energy of satellite
𝐸 = −
𝐺𝑀𝑚
2𝑟
Gravitational
potential
𝑉 =
𝑊
𝑚
=
𝐹∆𝑟
𝑚
Escape velocity
𝑉𝑒𝑠𝑐 =
2𝐺𝑀
𝑅
= 2𝑔𝑅
Newton’s law of
gravity
𝐹 =
𝐺𝑀𝑚
𝑟2 =
𝑚∆𝑉
∆𝑟
Centripetal force F=
𝑚𝑣2
𝑟
= 𝑚ω2𝑟

15. SciDoc
Questions
Q1. By using equations for potential and kinetic energy,
derive the equation for escape velocity
R
GM
v
2

Q2. Calculate the escape velocity for the following planets:
a) Mars: mass = 6.46 × 1023 kg, radius = 3.39 × 106 m
b) Mercury: mass = 3.35 × 1023 kg, radius = 2.44 × 106 m
c) Venus: mass = 4.90 × 1023 kg, radius = 6.06 × 106 m

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Answers

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Exam question
Complete the exam
questions.