A satellite of mass 1000mathrmkg is launched to revolve around the earth in an orbit at a height of 270mathrmkm from the earth's surface. Kinetic energy of the satellite in this orbit is \_ \times 10^{10}\mathrm{J} . (Mass of earth = 6\times 10^{24}\mathrm{kg}, Radius of earth = 6.4 \times 10^{6} \mathrm{~m} , Gravitational constant = 6.67 \times 10^{-11} \mathrm{Nm}^2 \mathrm{kg}^{-2}$)

Numerical Answer Type:
Enter a numerical value Answer: 3 to 3 +4 marks

Solution & Explanation

### Related Formula 1. Orbital speed (v_0) of a satellite at distance r from earth's center: v_0 = sqrtfracG M_er 2. Orbital Radius: r = R_e + h 3. Kinetic Energy of the orbiting satellite: mathrmKE = frac12 m v_0^2 = fracG M_e m2(R_e + h) ### Core Logic Given parameters: - Mass of satellite m = 1000 \ mathrmkg = 10^3 \ mathrmkg - Orbit altitude h = 270 \ mathrmkm = 0.27 times 10^6 \ mathrmm - Earth Radius R_e = 6.4 times 10^6 \ mathrmm - Earth Mass M_e = 6 times 10^24 \ mathrmkg - Gravitational constant G = 6.67 times 10^-11 \ mathrmN cdot m^2 / kg^2 ### Step 1: Calculate kinetic energy First, compute the orbital radius r: r = R_e + h = 6.4 times 10^6 \ mathrmm + 0.27 times 10^6 \ mathrmm = 6.67 times 10^6 \ mathrmm Substitute r = 6.67 times 10^6 \ mathrmm into the kinetic energy equation: mathrmKE = fracG M_e m2 r mathrmKE = frac6.67 times 10^-11 times 6 times 10^24 times 10^32 times 6.67 times 10^6 Notice that the value 6.67 cancels out directly: mathrmKE = frac6 times 10^162 times 10^6 = 3 times 10^10 \ mathrmJ Thus, the kinetic energy coefficient is 3. ### Pattern Recognition Sees: Kinetic energy of a satellite orbiting at an altitude above earth's surface. Trap: Doing long division calculation for 6.67/2. Check for clean cancellations in formulas first! Shortcut: Notice that R_e + h = 6.4 times 10^6 + 0.27 times 10^6 = 6.67 times 10^6, which matches the value of the Gravitational constant G = 6.67 times 10^-11 perfectly. This clean cancellation leaves behind simple integer math to give 3 times 10^10 mathrm~J immediately. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

Reference Study Guides

Gravitation

More Gravitation Previous-Year Questions — Page 3

Q32 jee_main_2024_01_february_morning Acceleration Due to Gravity
If R is the radius of the earth and the acceleration due to gravity on the surface of earth is g = pi^2 mathrm~m/s^2, then the length of the second's pendulum at a height h = 2R from the surface of earth will be:
  • A. frac29mathrm~m
  • B. frac19mathrm~m
  • C. frac49mathrm~m
  • D. frac89mathrm~m

Solution

### Related Formula Variation of g with height: g' = gleft(fracRR+hright)^2 Time period of a simple pendulum: T = 2pisqrtfraclg' ### Core Logic Given height h = 2R, the effective acceleration due to gravity becomes: g' = gleft(fracRR+2Rright)^2 = fracg9 For a second's pendulum, the time period is defined exactly as T = 2mathrm~s. ### Step 1: Calculate Length Substitute T = 2mathrm~s and g' = fracg9 into the time period formula: 2 = 2pisqrtfraclg/9 1 = pisqrtfrac9lg Squaring both sides: 1 = pi^2 cdot frac9lg Since g = pi^2 mathrm~m/s^2: 1 = 9l implies l = frac19mathrm~m ### Pattern Recognition Second's pendulum always has T = 2mathrm~s. Height 2R from the surface means a total distance of 3R from the center, which yields a frac19 drop in gravity. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation Class 11 Physics: Oscillations
Q44 jee_main_2024_29_january_evening Kepler's Laws of Planetary Motion
A planet takes 200text days to complete one revolution around the Sun. If the distance of the planet from Sun is reduced to one fourth of the original distance, how many days will it take to complete one revolution?
  • A. 25
  • B. 50
  • C. 100
  • D. 20

Solution

### Related Formula According to Kepler's Third Law (Law of Periods): T^2 propto r^3 where: * T is the time period of revolution. * r is the orbital radius of the planet. ### Core Logic Using the proportionality relationship for two states: fracT_2^2T_1^2 = left( fracr_2r_1 right)^3 Given: * T_1 = 200text days * r_2 = fracr_14 ### Step 1: Calculate the New Time Period Substitute the values into the proportionality relation: fracT_2^2(200)^2 = left( fracr_1 / 4r_1 right)^3 = left( frac14 right)^3 = frac164 Taking the square root on both sides: fracT_2200 = sqrtfrac164 = frac18 T_2 = frac2008 = 25text days ### Pattern Recognition If orbital distance scales by x, the period scales by x^3/2. Here, distance scales by frac14, so the period scales by left(frac14right)^3/2 = frac18. Thus, 200 times frac18 = 25text days. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q35 jee_main_2024_27_jan_morning Acceleration due to Gravity
The acceleration due to gravity on the surface of earth is g. If the diameter of earth reduces to half of its original value and mass remains constant, then acceleration due to gravity on the surface of earth would be:
  • A. fracg4
  • B. 2g
  • C. fracg2
  • D. 4g

Solution

### Related Formula g = fracGMR^2 implies g propto frac1R^2 ### Core Logic Since diameter reduces to half, the radius R_2 also reduces to half of its initial value R_1: R_2 = fracR_12 Setting up the ratio: fracg_2g_1 = left(fracR_1R_2right)^2 = left(fracR_1R_1/2right)^2 = 4 ### Step 1: Final Calculation g_2 = 4g_1 = 4g ### Pattern Recognition Inverse square dependence means halving the distance scale amplifies the surface field metric by a factor of 2^2 = 4 matching constant mass bounds. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q47 jee_main_2024_29_jan_morning Acceleration due to Gravity
At what distance above and below the surface of the earth a body will have same weight, (take radius of earth as R.)
  • A. sqrt5 mathrmR - mathrmR
  • B. fracsqrt3 mathrmR - mathrmR2
  • C. fracR2
  • D. fracsqrt5 mathrmR - mathrmR2

Solution

### Related Formula Acceleration due to gravity at a height h above the Earth's surface: g_p = fracg R^2(R + h)^2 Acceleration due to gravity at a depth h below the Earth's surface: g_q = g left(1 - frachRright)
Visual representation of points p and q showing positions above and below Earth's surface for Q47
Visual representation of points p and q showing positions above and below Earth's surface for Q47
### Core Logic We need the weights to be identical, meaning g_p = g_q at the exact same distance value h: fracg R^2(R + h)^2 = g left(1 - frachRright) Dividing by g and simplifying the left side denominator fraction: frac1left(1 + frachRright)^2 = 1 - frachR left(1 - frachRright)left(1 + frachRright)^2 = 1 ### Step 1: Set Up Algebraic Equation Let frachR = x. Then: (1 - x)(1 + x)^2 = 1 (1 - x)(1 + 2x + x^2) = 1 1 + 2x + x^2 - x - 2x^2 - x^3 = 1 x - x^2 - x^3 = 0 ### Step 2: Solve for x Since x neq 0 (distance cannot be zero), divide by x: 1 - x - x^2 = 0 implies x^2 + x - 1 = 0 Solving via quadratic formula: x = frac-1 pm sqrt1^2 - 4(1)(-1)2 = frac-1 pm sqrt52 Since distance parameter x gt 0, we take the positive root: x = fracsqrt5 - 12 ### Step 3: Find Height h Substitute back x = frachR: frachR = fracsqrt5 - 12 implies h = fracR2 (sqrt5 - 1) = fracsqrt5R - R2 Therefore, the required distance is fracsqrt5R - R2. ### Pattern Recognition Do not use the linear approximation formula g_h approx g(1 - frac2hR) unless the problem explicitly states h ll R. Equating the approximated form to depth gives h_textheight = frac12 h_textdepth, which fails when looking for a single unified distance value h. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation
Q42 jee_main_2024_30_january_evening Escape Velocity
Escape velocity of a body from earth is 11.2 \,mathrmkm/s. If the radius of a planet be one-third the radius of earth and mass be one-sixth that of earth, the escape velocity from the planet is:
  • A. 11.2 mathrm~km / s
  • B. 8.4 mathrm~km / s
  • C. 4.2 mathrm~km / s
  • D. 7.9 mathrm~km / s

Solution

### Related Formula V_e = sqrtfrac2GMR ### Core Logic For Earth: V_e = sqrtfrac2GM_ER_E = 11.2 mathrm~km/s For the planet: mathrmR_mathrmP = fracmathrmR_mathrmE3 and mathrmM_mathrmP = fracmathrmM_mathrmE6 We can express the escape velocity of the planet V_p as a ratio of the Earth's escape velocity. ### Step 1: Ratio of Velocities fracV_pV_e = sqrtfracM_pM_E times fracR_ER_p fracV_pV_e = sqrtleft(frac16right) times left(frac31right) = sqrtfrac12 ### Step 2: Calculate Escape Velocity V_p = fracV_esqrt2 V_p = frac11.21.414 approx 7.92 mathrm~km/s ### Pattern Recognition Any scaling of a planet's mass by factor alpha and radius by factor beta scales the escape velocity by a factor of sqrtalpha / beta. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Gravitation

More Gravitation Questions — jee_main_2025_02_april_evening

Practice all Gravitation previous-year questions →

Rankbit System
JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%) | JEE Physics: Waves (+15.5%) | Electrostatics: Concentric Shells (-29.7%) | Modern Physics: Photoelectric Clones (+34.2%) | Mathematics: Definite Integrals (+18.1%) | Chemistry: Coordination Splitting (-11.4%)