Q#7
A satellite is projected vertically upwards from an earth station. At what height above the earth's surface will the force on the satellite due to the earth be reduced to half its value at the earth station? (Radius of the earth is 6400 km.)Answer:
Let the mass of the satellite be m kg. Force on the satellite due to earth at the earth station =mg, where g is the acceleration due to gravity.
Given Radius of the earth R = 6400 km = 64 x 105 m
From the gravity equation, we have
Mg = GMm/R²
(G= Universal constant of gravity, M= mass of the earth)
Let at a distance of r from the earth's surface this force reduces to half that is mg/2, Now distance of the satellite from the center of the earth = R + r, Now the equation for this situation is
mg/2 = GMm/(R + r)²
mg = 2GMm/(R + r)²
Equating the RHS expressions of equations (i) and (ii) we get,
GMm/R² = 2GMm/(R + r)²
(R + r)² = 2R²
R + r = R√2
r = (√2 – 1)R = 0.414 x 64 x 105 m ≈ 2650 km.
Q#8
Two charged particles placed at a separation of 20 cm exert 20 N of Coulomb force on each other. What will be the force if the separation is increased to 25 cm?
Answer:
Coulomb force F = (9 x 109)q1q2/r² N
9 x 109 q1q2 = Fr² = 20 x (0.20)² = 0.80 Nm² {r = 20 cm = 0.20 m}
When separation is changed to r = 25 cm = 0.25 m
Force = 9 x 109q1q2/(0.25)² = 0.80/(0.25)² = 0.80x 16 = 12.8 N ≈ 13 N
Alternately:
Since the force is inversely proportional to the square of the distance between the charges, the two forces F1, F2 and the separations r1, r2 can be expressed as
F1/F2 = (r2/r1)²
F2 = (r1/r2)²F1 = (0.20/0.25)² x 20 N = (4/5)² x 20 N = 12.8 N ≈ 13N
Q#9
The force with which the earth attracts an object is called the weight of the object. Calculate the weight of the moon from the following data:
The universal constant of gravitation G = 6.67 x 10-11 Nm²/kg², the mass of the moon = 7.36 x 1022 kg, the mass of the earth 6x 1024 kg and the distance between the earth and the moon = 3.8 x 105 km.
Answer:
We have M1 = 6 x 1024 kg , M2 = 7.36 x 1022 kg ,
r = 3.8 x 105 km = 3.8 x 108 m, G = 6.67 x 10-11 Nm²/kg²,
Now the weight of the moon = GM1M2/r²
= 6.67 x 10-11 x 6 x 1024 x 7.36 x 1022/(3.8 x 108)²
= 20.4 x 1019 ≈ 2 x 1020 N
Q#10
Find the ratio of the magnitude of the electric force to the gravitational force acting between two protons.
Answer:
Let the separation between protons be 'r'.
Electric force F = 9 x 109q2/r² N {where q is the charge on each proton}
Gravitational force P = Gm²/r² N {Where m is the mass of each proton and G is the universal constant of gravity}
We have q = 1.6 x 10-19 C, m = 1.672 × 10-27 kg, G = 6.67 x 10-11 Nm²/kg²
So the required ratio F/P = 9 x 109 q2/Gm²
= 9 x 109(1.6 x 10-19)²/[6.67 x 10-11 (1.672 × 10-27)²]
= 1.235 x 1036 ≈ 1.24 x 1036
(So it is clear that gravitational force between two protons is nothing in comparison to electric force)
Q#11
The average separation between the proton and the electron in a hydrogen atom in the ground state is 5.3 x 10-11m. (a) Calculate the Coulomb force between them at this separation. (b) When the atom goes into its first excited state the average separation between the proton and the electron increases to four times its value in the ground state. What is the Coulomb force in this state?
Answer:
(a) Charge on each of them = q = 1.6 x 10-19 C
Separation between them = r1 = 5.3 x 10-11 m
So coulomb force between them
F = 9 x 109 q2/r1² N = 9 x 109(1.6 x 10-19/5.3 x 10-11)² N
F = 8.2 x10-8 N
(b) Since the Coulomb force is inversely proportional to the square of the separation between the charges, Hence when the separation increases four times the force decreases sixteen times. So in the excited state, the force will be
= (8.2/16) x 10-8 N = 5.1 x 10-9
Q#12
The geostationary orbit of the earth is at a distance of 36000 km from the earth's surface. Find the weight of a 120-kg equipment placed in a geostationary satellite. The radius of the earth is 6400 km.
Answer:
The weight of the equipment is the gravitational force between it and the earth which is balanced by the centrifugal force on the equipment due to its revolution around the earth. In case of a geostationary orbit, the time of revolution is exactly equal to 24 hrs.
So, if we calculate the centrifugal force on the equipment, it will give its weight.
Hence weight = Centrifugal force = mω²r
We have m =120 kg, r = 6400 + 36000 km = 42400 km = 4.24 x 107 m,
Time period T = 24 h = 24 x 3600 s = 86400 s
ω= 2π/T= 2π/86400 rad/s
Now weight = 120 x (2π/86400)² x 4.24 x 107 N
= 26.88 N ≈ 27 N
We can solve it in another way too, as follows:
Weight of equipment on the surface of the earth = mg = 120 x9.8 N
= 1176 N {Taking g = 9.8 m/s² at the surface of the earth}
We know that the gravitational force is inversely proportional to the square of the distance. So if the distance is increased n times, the force will be reduced by n² times. Here this ratio
n = 42400/6400 = 6.625
So, the weight of the equipment in the satellite
= 1176/n²
= 1176/6.625² N ≈27 N
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