KKMishra's Tutorials: Solutions to Problems on "GRAVITATION" - H C Verma's Concepts of Physics, Part-I, Chapter-11, EXERCISES, Q_31 to Q

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GRAVITATION:--
EXERCISES- Q-31_to_Q39 and Extra 40th Problem

31. A Mars satellite moving in an orbit of radius 9.4x10³ km takes 27540 s to complete one revolution. Calculate the mass of Mars.

ANSWER: Radius of Mars, a = 9.4x10³ km =9.4x10⁶ m.
The time period, T =27540 s. Let mass of Mars = M then,
T² =4π²a³/GM
→M = 4π²a³/GT²
→M = 4*3.14*3.14*(9.4x10⁶)³/6.67x10⁻¹¹*27540*27540
= 32757x10¹⁸⁺¹¹/5.06x10⁹
=6.5x10³x10²⁹⁻⁹
=6.5x10²³ kg

32. A satellite of mass 1000 kg is supposed to orbit the earth at a height of 2000 km above the earth's surface. Find (a) its speed in the orbit, (b) its kinetic energy, (c) the potential energy of the earth - satellite system and (d) its time period. Mass of the earth = 6x10²⁴ kg.

ANSWER: (a) The speed of a satellite is given by v =√(GM/a)
Here a = 2000+6400 km = 8.4x10⁶ m
∴ v = √{6.67x10⁻¹¹*6x10²⁴/8.4x10⁶}
=√4.76x10²⁴⁻¹¹⁻⁶
=√4.76x10⁷
=6899 m/s
≈6.90 km

(b) The kinetic energy of the satellite =½mv²
=½*1000*(6900)² J
=(69*69/2)x10⁷ J
=2380x10⁷ J
=2.38x10¹⁰ J

(c) The potential energy of the earth satellite system with reference to infinity as zero P.E. = -GMm/a
=-6.67x10⁻¹¹*6x10²⁴*1000/8.4x10⁶
=-4.76x10⁻¹¹⁺²⁴⁺³⁻⁶ J
=-4.76x10¹⁰ J

(d) The time period T of a satellite is given by
T² =4π²a³/GM
=4*3.14*3.14*(8.4x10⁶)³/6.67x10⁻¹¹*6x10²⁴
=584x10¹⁸⁺¹¹⁻²⁴
=584x10⁵
→T = 7642 s = 7642/3600 h = 2.12 hours

33. (a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of earth's rotation. (b) If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth = 6x10²⁴ kg.

ANSWER: (a) Since the angular speed of the satellite and the earth is same both will take the same time to complete one revolution. Therefore the time period of the satellite T = 24 hours
=24*3600 s
If the radius of the circular orbit = a, then
T² =4π²a³/GM
→a³ = T²GM/4π²
=24*3600*24*3600*6.67x10⁻¹¹*6x10²⁴/4π²
=6*36*24*36*6.67*6x10²⁴⁻¹¹⁺⁴/π²
=7.469x10⁶*10¹⁷/π²
=0.7568x10²³
=0.07568x10²⁴
→a = 0.42335x10⁸ m
=0.42335x10⁵ km
=42335 km
≈42300 km

(b) As it is clear from the diagram below the satellite revolves ¼th of a complete revolution in going from above the North pole to over the equatorial plane.

Diagram for Q-33

Since the time taken in one complete revolution is 24 hours. The time taken to come over the equatorial plane =¼*24 hours
= 6 hours.

34. What is the true weight of an object in a geostationary satellite that weighed exactly 10.0 N at the north pole?

ANSWER: Since the weight of an object is the force applied by the earth on it and it is inversely proportional to the square of the distance from the center of the earth. If the radius of the earth is R and the distance of the object from the surface of the earth is h then
The weight at the north pole W ∝ 1/R²
and the weight in the satellite W' ∝ 1/(R+h)²
∴ W'/W =R²/(R+h)²
→W' = WR²/(R+h)²
We have W = 10.0 N (given)
R = 6400 km = 6.4x10⁶ m
R+h = 6400+36000 = 42400 km =4.24x10⁷ m
∴W' = 10*(6.4x10⁶)²/(4.24x10⁷)² N
=(6.4*6.4/4.24*4.24)x10¹³⁻¹⁴ N
=2.28x10⁻¹ N
=0.228 N ≈0.23 N

35. The radius of a planet is R₁ and a satellite revolves around it in a circle of radius R₂. The time period of revolution is T. Find the acceleration due to the gravitation of the planet at its surface.

ANSWER: The time period is given by T² =4π²R₂³/GM
→GM = 4π²R₂³/T²

The acceleration due to gravity is given by g =GM/R₁²
→g = (4π²R₂³/T²)/R₁²
→g = 4π²R₂³/T²R₁²

36. Find the minimum colatitude which can directly receive a signal from a geostationary satellite.

ANSWER: The geostationary satellite will be situated on the equatorial plane at about 36000 km above the surface of the earth and the signals from it will travel in straight lines. Due to the spherical shape of the earth, the signals cannot reach beyond the points it is tangential to the surface towards the poles. The latitude of these points = ∠AOB {See diagram below}
It is the maximum latitude where the signals can reach. Since the colatitude is the complementary angle. ∠COB =α is the minimum complementary angle where the signals can reach.

Diagram for Q-36

In the diagram ⧍AOC is a right-angled triangle with ∠AOC =90°. The radius OB will be perpendicular to the tangent AB. So ⧍AOB is also a right-angled triangle with ∠ABO =90°. From geometry,
∠OAB =∠COB =α = Colatitude
In triangle AOB,
sinα = OB/OA =R/(R+h) =6400 km/(6400+36000) km
→sinα =6400/42400 =0.15
→α = sin⁻¹(0.15)

37. A particle is fired vertically upward from earth's surface and it goes up to a maximum height of 6400 km. Find the initial speed of the particle.

ANSWER: At the earth's surface P.E. =-GMm/R
If the speed of the particle =v, then its K.E. =½mv²
Total Energy =K.E. + P.E. =½mv²- GMm/R
At the maximum height K.E. =0
And P.E. = -GMm/(R+h)
Total energy at Maximum height = -GMm/(R+h)
Equating,
½mv²- GMm/R = -GMm/(R+h)
→v² =2GM{1/R-1(R+h)}
=2GMh/(R+h)R
Here given that h=6400 km =R
∴v² =2GMR/2R² =GM/R =6.6x10⁻¹¹*6x10²⁴/6400x10³ =0.62x10⁻¹¹⁺²⁴⁻⁵
=0.62x10⁸ =
→v = 0.787x10⁴ m/s =7870 m/s =7.87 km/s ≈7.9 km/s

38. A particle is fired vertically upward with a speed of 15 km/s. With what speed will it move in interstellar space. Assume only earth's gravitational field.

ANSWER: v = 15 km/s =15000 m/s
Let the interstellar speed of the particle = v'
So its K.E. =½mv'² but here its P.E. = 0
Total energy = ½mv'²
At the earth's surface, total energy = K.E.+P.E.
=½mv²-GMm/R
The total energy will remain constant. Equating
½mv'² =½mv²-GMm/R
→v'² = v² - 2GM/R
=(15000)² - 2*6.67x10⁻¹¹*6x10²⁴/6400x10³
=2.25x10⁸ - 1.25x10⁻¹¹⁺²⁴⁻⁵
=2.25x10⁸ - 1.25x10⁸
=1.0x10⁸
→v' = √(1.0x10⁸) m/s
=1.0x10⁴ m/s =10 km/s

39. A mass of 6x10²⁴ kg (equal to the mass of the earth) is to be compressed in a sphere in such a way that the escape velocity from its surface is 3x10⁸ m/s. What should be the radius of the sphere?

ANSWER: Let the radius of the sphere be R. Given that
Mass M = 6x10²⁴ kg, Escape velocity v = 3x10⁸ m/s, then
v = √(2GM/R)
→R = 2GM/v² =2*6.67x10⁻¹¹x6x10²⁴/9x10¹⁶
→R =8.89x10²⁴⁻¹¹⁻¹⁶ =8.89x10⁻³ m = 8.89x10⁻³x10³ mm
→R = 8.89 mm ≈ 9 mm

The below question is out of Book ↓
40. A space station will revolve around the earth in a circular orbit at some unknown height. The astronauts find it difficult to work in weightlessness for a long time. To solve this problem the engineers want to design the working place in the space station in a shape of a cylinder that will rotate about its axis to give a centrifugal force on its surface equal to the weight of a body on earth's surface. Find the required time period of the cylindrical working place if its diameter is 16 m.

ANSWER: Radius of the workplace, r = 16/2 = 8 m
If the speed of an object of mass m at the surface of the workplace = v,
The centrifugal force on the object = mv²/r
This should be equal to the weight of the object on the surface of the earth = mg.
Equating these two,
mv²/r = mg
→v² = gr = 9.80*8 = 78.4
→v =√78.4 =8.85 m/s
The required time period of the rotation T = 2πr/v
→T = 2*3.14*8/8.85 s
→T = 5.68 s

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