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

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Gravitation

More Gravitation Previous-Year Questions — Page 2

Q14 jee_main_2025_04_april_morning Escape Velocity and Potential Energy
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: The kinetic energy needed to project a body of mass m from earth surface to infinity is frac12mgR, where R is the radius of earth. Reason R: The maximum potential energy of a body is zero when it is projected to infinity from earth surface. In the light of the above statements, choose the correct answer from the option given below
  • A. A is False but R is true
  • B. Both A and R are true and R is the correct explanation of A
  • C. A is true but R is false
  • D. Both A and R are true but R is NOT the correct explanation of A

Solution

### Related Formula Escape kinetic energy requirement formulation: textKEtextescape = fracG M mR = m g R Gravitational potential energy field definition: U = -fracG M mr At r to infty, Utextmax = 0. ### Core Logic * **Assertion Check:** The work required to move an object from R to infty is Delta U = 0 - left(-fracGMmR ight) = fracGMmR = mgR. The Assertion states it is frac12mgR, which represents only the orbital velocity metric or binding equivalent. Thus, Assertion A is **false**. * **Reason Check:** Since gravitational energy is attractive and negative, it strictly ascends asymptotically up to a ceiling limit value of zero at infinity. Hence, Reason R is **true**. ### Pattern Recognition Escape energy = mgR. Binding energy magnitude = frac12mgR (for circular orbiters). Zero represents the reference point ceiling of a negative bound system state. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q6 jee_main_2025_07_april_evening Kepler's Laws of Planetary Motion
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R). Assertion (A): The radius vector from the Sun to a planet sweeps out equal areas in equal intervals of time and thus areal velocity of planet is constant. [cite: 55] Reason (R) For a central force field the angular momentum is a constant. [cite: 56] In the light of the above statements, choose the most appropriate answer from the options given below: [cite: 57]
  • A. Both (A) and (R) are correct and (R) is the correct explanation of (A) [cite: 63]
  • B. Both (A) and (R) are correct but (R) is not the correct explanation of (A) [cite: 65]
  • C. A is correct but R is not correct [cite: 72]
  • D. A is not correct but R is correct [cite: 73]

Solution

### Related Formula fracdAdt = fracL2m [cite: 684] ### Core Logic Kepler's Second Law state that the areal velocity fracdAdt is directly proportional to the angular momentum L of the planet[cite: 55, 684]. Because the gravitational force between the Sun and the planet acts strictly along the line joining their centers (a central force field), its torque vectau = vecr times vecF = 0[cite: 56, 686]. Since torque is zero, the angular momentum L remains completely constant over time[cite: 686]. As a consequence, fracdAdt = textconstant[cite: 55, 684]. Both statements are true and (R) is the correct explanation of (A)[cite: 63]. ### Pattern Recognition Areal velocity constancy is a direct geometric manifestation of the conservation of angular momentum under any central force field[cite: 55, 56, 686]. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q25 jee_main_2025_24_jan_evening Acceleration due to Gravity
Acceleration due to gravity on the surface of earth is 'g'. If the diameter of earth is reduced to one third of its original value and mass remains unchanged, then the acceleration due to gravity on the surface of the earth is ____ g.
Numerical Answer. Answer: 9 to 9

Solution

### Related Formula Surface gravitational acceleration: g = fracGMR^2 ### Core Logic Since diameter drops to 1/3, the radius R' also scales down to 1/3 its original value (R' = R/3), while mass M remains constant. New acceleration value g' calculation: g' = fracGM(R/3)^2 = 9 cdot fracGMR^2 = 9g The scaling factor is 9. ### Pattern Recognition Gravity follows an inverse-square law with respect to radius. Shrinking the radius by a factor of n increases surface gravity by n^2 if the mass is unchanged. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q14 jee_main_2025_24_jan_morning Kepler's Laws of Planetary Motion
A satellite is launched into a circular orbit of radius 'R' around the earth. A second satellite is launched into an orbit of radius 1.03 R. The time period of revolution of the second satellite is larger than the first one approximately by :-
  • A. 3%
  • B. 4.5%
  • C. 9%
  • D. 2.5%

Solution

### Related Formula By Kepler's Third Law of Planetary Motion, the square of the orbital period T is proportional to the cube of the orbital radius R: T^2 = K cdot R^3 ### Core Logic Taking logs and differentiating to find fractional errors for small changes: 2 fracDelta TT = 3 fracDelta RR fracDelta TT = frac32 left(fracDelta RR ight) ### Step 1: Computing Percentage Change The change in radius is Delta R = 1.03R - R = 0.03R, which means fracDelta RR = 0.03 or 3\% [cite: 106, 708]. Substitute this into our fraction scaling relation: fracDelta TT = frac32 times 0.03 = 0.045 = 4.5\% ### Pattern Recognition Power factors act as direct linear multipliers for small percentage shifts. Here, the scaling factor is simply frac32 times the radius change. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q11 jee_main_2025_28_jan_evening Escape Velocity
Earth has mass 8 \times and radius 2 \times that of a planet. If the escape velocity from the earth is 11.2 \, textkm/s , the escape velocity in textkm/s from the planet will be:
  • A. 11.2
  • B. 5.6
  • C. 2.8
  • D. 8.4

Solution

### Related Formula The expression for escape velocity from a spherical planetary body is given by: v_textescape = sqrtfrac2GMR ### Core Logic Let the planet's mass be M_P and its radius be R_P. According to the problem statement : * Earth's mass, M_E = 8 M_P implies fracM_PM_E = frac18 * Earth's radius, R_E = 2 R_P implies fracR_ER_P = 2 Taking the ratio of escape velocities : fracv_Pv_E = sqrtleft(fracM_PM_Eright) times left(fracR_ER_Pright) fracv_Pv_E = sqrtfrac18 times 2 = sqrtfrac14 = frac12 Given that v_E = 11.2 text km/s: v_P = frac12 times 11.2 = 5.6 text km/s ### Pattern Recognition Setting up quick ratios prevents substitution mistakes. For any planetary variant, notice how scaling properties scale inside the root operator directly. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

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