Density Calculations

Kepler found three laws which governed planetary motion. One linked their orbital periods and their distance from the sun. The orbital periods can be found by painstaking observation of the motions of the planets in the sky.  Kepler’s 2^nd Law can then be used to find the radius of orbit. Once this is known Newton’s Law of gravitation can be used to calculate the mass of a planet. The diameter of a planet can then be calculated form the angle it subtends when viewed from Earth. 

Name	Radius km	Mass kg
Sun	695000	1.99x1030
Mercury	2440	3.3x1023
Venus	6052	4.87x1024
Earth	6378	5.97x1024
Mars	3397	6.42x1023
Jupiter	71492	1.90x1027
Saturn	60268	5.68x1026
Uranus	25559	8.68x1025
Neptune	24766	1.02x1026
Pluto	1137	1.027x1022
Moon	1738	7.35x1022
Ganymede	2634	1.48x1023

1.     Using the table opposite calculate the average densities of the objects listed.

2.     The planets can be divided into two groups called Jovian and Terrestrial. Identify which belongs to each group, and why they are so named.

3.     The average density of the surface rocks of the earth is 2 800kg m^-3.  How does this compare with your value. The earth is thought to have a dense iron core. Why do you think this conclusion has been drawn?

4.       Comment upon my use of the term average density in question 1.

5.     Compare your values for Mars and the Earth.  The surface rocks on Mars are similar to igneous rocks on Earth. Mars is thought to have a core rich in sulphur as well as iron, explain why.

6.     The Moon and the Earth are the same age. This has been verified from rocks returned by American and Soviet missions. Compare the densities of the two bodies.

7.     There are four theories of the formation of the Earth Moon system

a)           That the two bodies formed together at the same time.

b)           That a proto-planet of the size of Mars collided with the proto-earth. Both planets were destroyed. The bulk of the material accreted to form the Earth but a small amount of lighter material spun out and formed the Moon.

c)            The teardrop theory where the proto moon was torn away from the Earth leaving a hole now filled by the Pacific Ocean. (Continental drift?)

d)           The moon was a stray planet that the earth captured. (compare to other small objects)

 

 

 

8.     The planet mercury superficially resembles the Moon. Do you think its internal structure is similar? Give reasons for your answer.

9.     The Jovan planets are sometimes called "Gas Giants". With reference to their densities give reasons why. Neptune and Uranus are both blue gas giants. Some may call them sister planets. Comment upon this.

10.  Should Pluto be classed as a terrestrial planet?

 

Moments question weight of retort stand

(a) Principle of moments

For equilibrium / balance

sum of moments clockwise = sum of moments anti-clockwise

or sum of moments (about a point) is zero

[Sum or equivalent eg total/net/resultant [NOT all] must be seen at least

once] 2

(b)(i) Weight of retort stand

Use of principle of moments

i.e.180 × (10-3 m) × W = 228 × (10-3 m) × 9 (N)

[Allow 1 wrong distance]

11.4 (N)

[At least 3 sig fig required. No u.e.] [In this question allow reverse

calculation to gain full marks] [Bald answer scores 0]

 2

(ii) Reading on newtonmeter and normal contact

Fully correct moment equation

F × 448 × (10-3 m) = 11.4 (N) ×40 × (10-3 m)

[allow 11 N × 40 × (10-3) m or ecf from bi]

= 1.0 N

Normal contact force (11.4 N/11 N − 1 N) = 10.4 N or 10 N (upwards)[ecf

Moments Weight of pen

a

Weight of pen

Weight = mg seen or used

[ignore power of 10 error]

Answer [ 0.11 N]

Eg Weight = 0.011 kg x 9.81 m s-2

= 0.108 N

2 marks

bi

Add labelled arrow to represent weight

Arrow pointing down at 8 cm labelled weight (of pen) /0.11 N /W/mg /their value

􀀹

[Check by eye. Arrow will be along the vertical line of the last letter ‘d’ in the word

‘balanced’ in the line above]

1 mark

bii Calculate weight of top(ecf their value of weight)

Use of principle of moments i.e. a clockwise moment equal to an anti-clockwise moment

[Give this mark even if distances are wrong, but must use 0.11 N / their value]

Correct distances used [must be 1 cm and 4cm, no ECF from bi]

 

Answer [ 0.03 N ]

Eg 0.11 N x 1 cm = W x 4 cm

W = 0.0275 N

[Only penalise same unit error once]

3 marks

Moments a painter walks the plank

Answers to sheet on moments

Moments

You must draw a diagram for each question with forces and distances from pivot (fulcrum) added. (g = 9.81 Nkg^-1)

62 A metre rule is pivoted at its mid-point. A 0.60 N weight is suspended from one end. How far from the other end must a 1.00 N weight be suspended for the rule to balance? 3.5N

63 A metre rule is pivoted at its mid-point and a 50 g mass suspended from the 20cm mark. What mass balances the rule when suspended from the 65 cm mark? 100g
64 A lever of negligible weight is 2 m long. If a 0.80 N weight at one end balances a 0.20 N weight at the other end, how far is the fulcrum from the 0.80 N weight? 40cm
65 A metre rule is pivoted at its mid-point. A 1.0 N weight is suspended from the 30cm mark and a 2.0 N weight from the 90cm mark. Where must an upward force of 3.0 N be applied to balance the rule? 70cm
66 A uniform lever 150 cm long is pivoted at its mid-point. A 50 g mass is suspended from the left-hand end and an 80 g mass from the right- hand end. A string tied 50 cm to the right of the fulcrum passes upwards over a pulley so that the string is at 30° to the lever. What mass must be suspended from the string for the lever to balance? 90g
67 With the help of a diagram explain how you would find the weight of a metre rule using only a fulcrum and a weight which can be suspended from the rule.

68 A uniform rod AB, 2 m long, weighs 0.40 N. If weights of 0.80 N and 0.40 N are suspended from A and B respectively, at what point will it balance? 75cm from A
69 A uniform metre rule weighs 1.20 N. Weights of 0.10 N, 0.50 N, 1.00 N and 0.20 N are suspended respectively from the 10 cm, 20 cm, 60cm and 80cm marks. At which mark will the rule balance? 49cm mark
70 A non uniform rugby post 6 m long lies flat on the ground. When the narrow end only is lifted off the ground, the force needed is 140 N. When the thick end only is lifted the force needed is 280 N. Deduce the weight of the post and the distance of the centre of gravity from the thick end. 420N 2m
71 A uniform bridge AB, 30 m long and weighing 200 000 N, rests on supports at each end. Find the forces on the supports when a car of weight 10000 N is 4 m from A and a lorry of weight 100 000 N is 10 m from B. 142 kN 168 kN
72 A scaffold plank AB, 5m long, weighs 200N and its centre of gravity is 2 m from A. It rests on two bars, one 0.5 m from A, the other 0.6 m from B. Find the weight of the heaviest person who can walk from end to end without tilting the plank. 600N
73 A metre rule is suspended from two balances calibrated in Newtons. One balance is attached at the 5 cm mark, the other at the 80cm mark. If the rule weighs 1 N and downward forces of 2 N are applied at the 10cm and 60cm marks, find the readings on the balances. 2.8N 2.2N

74 A metre rule is pivoted freely at its mid-point. Equal, opposite and parallel vertical forces are applied at each end. If the direction of these forces ALWAYS remains vertical, what will happen? Draw sketches to clarify your explanation.
75 A tapered rod of mass 200 g is 160 cm long. It balances at its midpoint when a 150 g mass hangs from the narrow end. How far is the centre of gravity from the thick end? 20cm
76 A metal tube of length 30 cm and weight 0.45 N is fitted with a wooden handle 10 cm long and weight 0.30 N. How far is the centre of gravity from the middle of the handle? 12cm

 

SHM