HCI 07A Gravitation Solutions (Self-Review Questions)

Gravitation Tutorial 7A Self Review Solution S1 C 2 2 22 2 31 2 2units of = units of = GMm FrFG Mmr Fr kg m s mGm k g s Mm kg       S2 Strategy: vector sum of  gravitational force acting  on a mass due to the  respective masses.   Strategy: always make the  physical quantity or constant that  you wish to find the subject first.   NOTE: pay attention to the  presentation of answers

S3 D 2 22 49(7 ) 49 GMmW R GMm GMm W RR   S4 tan 𝜃 ൌ ହ.଴ ଵଶ.଴ , θ = 22.6˚ 𝑟ൌ ඥ5.0ଶ ൅1 2 . 0ଶ ൌ 13.0 cm Field strength due to mass A, gA = field strength due to mass B, gB = ீሺଶ଴.଴ሻ ሺଵଷ.଴ൈଵ଴షమሻమ. At P, the vertical components gA cancels the vertical component of gB. Hence the resultant g is towards the left and its magnitude given by 𝑔ோ ൌ2 ቀ ீሺଶ଴.଴ሻ ሺଵଷ.଴ൈଵ଴షమሻమቁ cos 22.6° ൌ 1.46 ൈ 10ି଻ N kg-1 S5 D 2 GMg r , thus 2 Mg r (since G is a constant) S6 S7 A 22 2 22 22 1 ' ' () () GMm GMmg g g RR R gr g r gg rh rh        Strategy: vector sum of  gravitational field strengths  acting at the point due to  the respective masses.   Besides resolving forces in  two perpendicular  directions, another way is  to draw a force vector  triangle and use sine rule  or cosine rule to obtain the  resultant.  r  θ 12.0 cm  5.0 cm 5.0 cm  P  A  B  gA  gB  The trickier questions often give the “height / altitude of the  mass above surface of the Earth”. When applying the formula  for gravitational force, you should always consider the distance  between the two masses i.e. (Earth radius + height).  Since mass = density x volume Another question similar to S3 involving the “height / altitude  of the mass above surface of the Earth”. When applying the  formula for gravitational field strength, you should always  consider the radius as (Earth radius + height).

S8 B A uniform gravitational field means that the gravitational fie ld strength/force is constant. S9 D S10 C The gravitational force exerted by the Earth on the moon is th e real force that provides the Moon’s centripetal acceleration (Newton’s Second Law). S11 C A geostationary orbit must have the same angular velocity, hen ce the same period, as the Earth's rotation on its axis, so as to appear stationary to a point on Earth. S12 D The object in the space capsule orbiting the Earth seems to be floating suggests no normal contact force N acting on the object, hence weight is the only force acting on it. Both the space capsule and object experience the same centripetal acceleration towards the centre of the Earth, which is similar to the analogy of a man in a lift both free falling. In this case, the man will also experie nce apparent weightlessness as he falls as both him and lift are accelerating downwards at g = 9.81 m s -2, hence no normal contact force acting on him. Similar to lecture example 3

S13