2. Each planet revolves
around the sun in
elliptical orbits, with the
sun at one focus of the
ellipse.
Explanation:
P is the planet and S is
the sun. a and b are semi-
minor axis respectively.
3. The radius vector, drawn from the sun to a planet,
sweeps out equal areas in equal times.
Explanation:
The areal velocity of the radius vector is constant. If the
planet moves from A to B, C to d, E to F in equal time
interval ∆t, then the areas ASB,CSD and ESF are equal.
When the planet is nearest to sun, its velocity is
maximum. When the planet is farthest from the sun, its
velocity is minimum.
4. The square of the period of revolution of the
planet from the sun, larger will be the period
of revolution. The planet Pluto is farthest to
sun. Its period is 248 years.
5. Statement:
Every particle of matter in
the universe attracts every
other particle with a force
which is directly
proportional to the
product of their masses
and inversely proportional
to the product of their
masses and inversely
proportional to the square
of the distance between
them.
6. Explanation:
Let m1 and m2 be the masses of two particles situated at a
distance r apart. The force of attraction between them is
given by
F α m1m2/r2 or F = Gm1m2/r2
Here G is universal constant, called the universal
gravitational constant.
7. The force of attraction of the earth on a body of mass m
placed on it = mg.
Here, g is the acceleration due to gravity.
Let M = mass of the earth
R = Radius of the earth
Gravitational force of attraction between a body of
mass m and earth = GMm/R2
GMm/R2 = mg
8. M = gR2/G
Density of earth = ρ =
M/V
ρ = (gR2/G)/(4/3πR3) =
3g/4πRG
9. Apparatus:
A small mirror strip RS is suspended from a torsion
head H by means of a quartz fibre W. From the two
ends of the mirror, two identical gold spheres A and B
suspended. The two spheres are arranged to be at
different levels. The arrangement is enclosed in a fixed
inner tubeT1.
10. In the outer co-axial
tube T2, two large lead
spheres C and D are
suspended. The centre
of C is in level with that
of A, the centre of D is
in level with that of B
and the distance AC =
BD.
11. The outer tube is rotated so that the larger spheres are
on the opposite sides of the smaller spheres.
Gravitational attraction between the two pairs of
masses produces a torque on the mirror strip RS. Due
to this torque, the mirror is deflected through an angle
ϴ. The deflection ϴ is measured by a scale and
telescope arrangement.
12. Let M = mass of the lead sphere,
M = mass of the gold sphere,
d = distance between the two spheres.
Force of attraction between spheres A and C is
F = GMm/d2
Force of attraction between spheres B and D is
F = GMm/d2
Hence two equal unlike parallel forces act at R and S.
13. Let l be the mirror strip RS.
The deflecting torque = (GMm/d2)xl
A restoring torque is set up in the suspension fibre.
Let C be the torque per unit twist. Then, for a
deflection ϴ, the restoring torque on the fibre = cϴ
14. When the system is in equillibrium, the two torques must balance each
other.
(GMm/d2)xl = cϴ
G = cϴd2/mMxl (1)
Using the arrangement on the quartz fibre and the mirror strip with
gold spheres as a torsion pendulum, the period T is found. Then,
T = 2π
𝐼
𝑐
. Here I = moment of inertia of the suspended system.
From this c can be calculated.
Substituting for c in Eq.(1), G is calculated.
15. The use of quartz fibre makes the apparatus very
sensitive and accurate.
By the scale and telescope arrangement, very small
deflections can be measured accurately.
The two pairs of spheres are suspended at different
levels. So cross-attraction between the pairs of spheres
is eliminated.
