Initially a satellite of 100 kg is in a circular orbit of radius 1.5R_E. This satellite can be moved to a circular orbit of radius 3R_E by supplying alpha times 10^6J of energy. The value of alpha is ____. (Take Radius of Earth R_E = 6 times 10^6text m and g = 10text m/s^2)

Solution & Explanation

### Related Formula E = frac-GM_E m2r Delta E = E_f - E_i g = fracGM_ER_E^2 implies GM_E = g R_E^2 ### Core Logic Energy of a satellite in a circular orbit is given as E = frac-GM_E m2r where r is the radius of the circular orbit. Required energy to be supplied Delta E = E_f - E_i: Delta E = left( frac-GM_E m2(3R_E) right) - left( frac-GM_E m2(1.5R_E) right) Delta E = frac-GM_E m6R_E + fracGM_E m3R_E = fracGM_E m6R_E ### Step 1: Evaluate Delta E Substitute GM_E = g R_E^2 into the expression: Delta E = frac(g R_E^2) m6R_E = frac16 m g R_E = frac16 times 100 times 10 times (6 times 10^6) = 1000 times 10^6text J Comparing with alpha times 10^6text J, we get alpha = 1000. ### Pattern Recognition Orbital transition energy is always Delta E = fracGMm2 (frac1r_i - frac1r_f). Never forget the factor of 2 in the denominator (which accounts for kinetic energy contribution in orbit). ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

Reference Study Guides

Gravitation

More Gravitation Previous-Year Questions — Page 4

Q42 jee_main_2024_30_jan_morning Gravitational Potential and Field
The gravitational potential at a point above the surface of earth is -5.12 times 10^7 mathrm~J / kg and the acceleration due to gravity at that point is 6.4 mathrm~m/s^2. Assume that the mean radius of earth to be 6400 mathrm~km. The height of this point above the earth's surface is:
  • A. 1600 mathrm\,km
  • B. 540 mathrm\,km
  • C. 1200 mathrm\,km
  • D. 1000 mathrm\,km

Solution

### Related Formula V = -fracGM_ER_E + h g' = fracGM_E(R_E + h)^2 ### Core Logic The gravitational potential (V) and acceleration due to gravity (g') at a distance r = R_E + h from the center of the earth can be related by dividing their magnitudes: |V| / g' = r. ### Step 1: Set Up Equations From the given data: -fracGM_ER_E + h = -5.12 times 10^7 quad dots (i) fracGM_E(R_E + h)^2 = 6.4 quad dots (ii) ### Step 2: Isolate Variable Divide equation (i) by (ii) (taking magnitudes): R_E + h = frac5.12 times 10^76.4 R_E + h = 0.8 times 10^7 mathrm~m = 8000 mathrm~km ### Step 3: Solve for h Given mean radius of the earth R_E = 6400 mathrm~km: 6400 + h = 8000 h = 1600 mathrm~km ### Pattern Recognition Always exploit the V/g = r relationship to extract distances cleanly without having to substitute large values for G or M_E. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q46 jee_main_2024_31_jan_evening Escape Velocity
The mass of the moon is 1/144 times the mass of a planet and its diameter 1/16 times the diameter of a planet. If the escape velocity on the planet is v, the escape velocity on the moon will be:
  • A. fracv3
  • B. fracv4
  • C. fracv12
  • D. fracv6

Solution

### Related Formula v_textescape = sqrtfrac2GMR ### Core Logic For the planet: v = sqrtfrac2GM_pR_p For the moon: M_m = fracM_p144 and R_m = fracR_p16. ### Step 1: Setup the Ratio v_m = sqrtfrac2G M_mR_m v_m = sqrtfrac2G left(fracM_p144right)left(fracR_p16right) v_m = sqrtfrac2G M_pR_p times frac16144 ### Step 2: Simplification v_m = sqrtfrac2G M_pR_p times sqrtfrac19 v_m = v times frac13 = fracv3 ### Pattern Recognition Escape velocity scales as sqrtM/R. If M scales by x and R scales by y, velocity scales by sqrtx/y. Here, sqrt(1/144)/(1/16) = sqrt16/144 = sqrt1/9 = 1/3. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q37 jee_main_2024_31_jan_morning Superposition Principle
Four identical particles of mass m are kept at the four corners of a square. If the gravitational force exerted on one of the masses by the other masses is left(frac2sqrt2 + 132right)fracmathrmGm^2mathrmL^2, the length of the sides of the square is
  • A. fracmathrmL2
  • B. 4 L
  • C. 3L
  • D. 2 L

Solution

### Related Formula F = fracG m_1 m_2r^2 ### Core Logic
Superposition Principle diagram for Q37 - JEE Main 2024 Morning
Superposition Principle diagram for Q37 - JEE Main 2024 Morning
Let the side length of the square be a. Considering one corner mass, it experiences forces from the adjacent two masses (distance a) and the diagonally opposite mass (distance sqrt2a). The forces from the two adjacent masses are at 90^circ to each other: F = fracGm^2a^2 The resultant of these two is sqrt2F = sqrt2 fracGm^2a^2, directed along the diagonal. ### Step 2: Total Force Equation The force from the diagonal mass is: F' = fracGm^2(sqrt2a)^2 = fracGm^22a^2 Total resultant force F_textnet = sqrt2F + F': F_textnet = sqrt2 fracGm^2a^2 + fracGm^22a^2 = fracGm^2a^2 left( sqrt2 + frac12 right) F_textnet = fracGm^2a^2 left( frac2sqrt2 + 12 right) Equating this to the given force value: left(frac2sqrt2 + 132right)fracGm^2L^2 = fracGm^2a^2 left( frac2sqrt2 + 12 right) frac132 L^2 = frac12 a^2 a^2 = 16 L^2 a = 4L ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

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