physics

PUNCAK CEMERLANG CENTRE FOR EXCELLENCE
BANDAR SRI DAMANSARA

FORM 4 PHYSICS
ONLINE CLASS

GRAVITATION PART 2

1) Look at the above picture . What is happening to the apple ?
It is ……………………… to the ground.

2) Why did the apple fall to the ground ?
There is a ………………. due to gravity pulling the apple to the ………………

3) What is this force called ?
…………………………….. force

4) What is the formula for this force ?
F = where m= mass of object in kg and g = acceleration gue to
gravity

5) Lets recollect what we learnt last week .
Let the mass of the apple = m kg
Let the mass of the Earth = M kg
Let r = distance between centre of apple
and centre of Earth.
Let F = Force of attraction between the
Earth towards the Apple .


Hence F =

But F which is gravitational force is
also = mg

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Gravitational force acting on the Force of attraction on the Apple
apple from the Earth

In physics, 'g' is the symbol used for “acceleration due to gravity” of earth on any object on
earth. If observed more closely, it's value is same as that of earth's gravitational field or we
can say that acceleration due to gravity 'g' is the gravitational field of earth acting on a
object

Discussion:

1. Write the equation relating g with G , M and r .

2. What are the factors that influence the value of gravitational acceleration, g on an
object ?

………………………………………………………………………………………………………
3. From the equation in (1) , what are the constant factors?

………………………………………………………………………………………………………
4. Which factor affects g ?

………………………………………………………………………………………………………

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5. Lets investigate further the change in the values of g with r.
6. Look at the diagram below .

7. Calculate the value of g for the five distances shown above using the above data .

a) At R ( Anywhere on the Earth )

= =GM

g = r2
6.67 x 10−11 x 5.97 x 1024 9.813 Nkg-1 / ms-2
6.37 x 106

b) At 2R

g = 3.9819 x 1014

Ans : 2.45 Nkg-1 / ms-2

c) At 3R

g = 3.9819 x 1014

d) At 4R Ans : 1.090 Nkg-1 / ms-2
Ans : 0.6133 Nkg-1 / ms-2
g = 3.9819 x 1014

e) At 5R

g = 3.9819 x 1014

Ans : 0.3925 Nkg-1 / ms-2

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Now answer the following questions :
1. What is the value of gravitational acceleration on the Earth’s surface?

…………………………………………………………………………………………………
2. Plot a graph of g against R.

3. How does the value of gravitational acceleration change when the distance from the
centre of the Earth increases?

…………………………………………………………………………………………………………
4 Discuss the condition where the value of gravitational acceleration is almost zero.
…………………………………………………………………………………………………………

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5 The Figure below shows a sketch of a graph with various values of gravitational
acceleration, g and distance, r from the centre of the Earth.

The Figure below shows a satellite at height, h from the surface of the Earth. R is the
radius of the Earth and r is the distance of the satellite from the centre of the Earth, which
is the radius of the orbit.

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Practice questions
1) A satellite orbits around the Earth at a height of 480 km. What is the value of
gravitational acceleration at the position of the satellite?
[ G = 6.67 × 10 –11 N m 2 kg –2, Mass of earth = 5.97 × 10 24 kg,
Radius of earth = 6.37 × 10 6 m ]
Ans : 8.49 ms-2

2) What is the gravitational field strength of a satellite at a point 900 km above the
earth’s surface?
[ G = 6.67 × 10 –11 N m 2 kg –2, Mass of earth = 5.97 × 10 24 kg,
Radius of earth = 6.37 × 10 6 m ]

Ans : 7.53 Nkg-1 / ms-2

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3) What is the gravitational field strength on the surface of the sun?
Radius of the Sun = 6.96 x 10 8 m
Mass of the Sun = 1.99 x 10 30 kg
G = 6.67 × 10 –11 N m 2 kg –2

Ans : 274 Nkg-1 / ms-2

4) Calculate the gravitational acceleration between the Earth and a geostationary
satellite , if we assume that it orbits at a height of 35,800 km above the Earth’s
surface.
G = 6.67 × 10 –11 N m 2 kg –2 , Radius of earth = 6.37 × 106 m
Mass of earth = 5.97 × 10 24 kg

Ans : 0.224 Nkg-1 / ms-2

5) The centre of the moon is about 384,400 km from the centre of the earth.
The Mass of the earth = 5.97 × 10 24 kg
G = 6.67 × 10 –11 N m 2 kg –2
Determine the gravitational field strength that the moon will experience

Ans : 0.0027 Nkg-1 / ms-2

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6) A satellite of mass 900kg orbits 400km above the Earth’s surface as shown below.

Calculate the gravitational field strength that the satellite experiences at this point.
G = 6.67 × 10 –11 N m 2 kg –2 , Radius of earth = 6.37 × 106 m
The Mass of the earth = 5.97 × 10 24 kg

Ans : 8.703 Nkg-1 / ms-2

7) Calculate the gravitational field strength on the surface of Mars.
Mars has a radius of 3.43 x10 6 m and a mass of 6.37 x 10 23 kg.
G = 6.67 × 10 –11 N m 2 kg –2

Ans : 3.611 Nkg-1 / ms-2

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8) At what distance from the Earth’s surface is the gravitational acceleration 7.33 ms-2 ?
G = 6.67 × 10 –11 N m 2 kg –2

The Mass of the earth = 5.97 × 10 24 kg

Ans : 7.37 x 10 6 Nkg-1 / ms-2

9) The Earth orbits around the Sun . The distance between the centre of the Earth and
the centre of the Sun is 1.46 x 10 10 m . What is the strength of the Sun’s
gravitational field at this distance?

Mass of the Sun = 1.99 x 10 30 kg
G = 6.67 × 10 –11 N m 2 kg –2

Ans : 0.622 Nkg-1 / ms-2

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10) A piece of space junk of mass 24 kg is at a distance of 7.00 x 10 6 m from the
centre of the Earth. What is the gravitational force between the space junk and the
Earth?
The Mass of the earth = 5.97 × 10 24 kg
G = 6.67 × 10 –11 N m 2 kg –2
Ans : 195.03 N

11) A weather satellite orbits the Earth at a height of 560 km. What is the value of
gravitational acceleration at the position of the satellite?
[ G = 6.67 × 10 –11 N m 2 kg –2, Mass of earth = 5.97 × 10 24 kg,
Radius of earth = 6.37 × 10 6 m ]

Ans : 8.31 Nkg-1 / ms-2

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