What is minimum energy required to launch a satellite of mass m from the surface of a planet of mass m and radius r in a circular orbit at an altitude 2R:
Correct answer: D. 5GmM/6R
- A. 2GmM/3R
- B. GmM/2R
- C. GmM/3R
- D. 5GmM/6R
Explanation
To solve this problem, we can use the concept of gravitational potential energy. The potential energy of an object due to gravity is given by:PE = -G * M * m / rwhere:G is the gravitational constantM is the mass of the planetm is the mass of the satelliter is the distance between the center of the planet and the satelliteIn this scenario, we need to consider the initial and final potential energies of the satellite:Initial potential energy (PE_i): When the satellite is at the surface of the planet (r = R), its potential energy is:PE_i = -G * M * m / RFinal potential energy (PE_f): When the satellite reaches the circular orbit at an altitude of 2R (r = 3R), its potential energy is:PE_f = -G * M * m / (3R)Escape velocity: The minimum energy required to launch the satellite and overcome the gravitational pull of the planet is known as the escape velocity (v_e). It can be calculated using the following equation:1/2 * m * v_e^2 = -G * M * m / RSolving for v_e, we get:v_e = sqrt( 2 * G * M / R )Kinetic energy (KE): Once the satellite reaches the circular orbit, it needs to have enough kinetic energy to maintain its orbital motion. This kinetic energy can be calculated using the following equation:KE = 1/2 * m * v^2where v is the orbital velocity of the satellite.Total energy (E): The minimum total energy required (E) is the sum of the escape velocity and the kinetic energy needed for the orbit:E = KE + PE_f = 1/2 * m * v^2 - G * M * m / (3R)Relating orbital velocity and escape velocity: In a circular orbit, the orbital velocity (v) is related to the escape velocity (v_e) by:v = v_e / sqrt(2)Substituting this relationship into the equation for total energy:E = 1/2 * m * (v_e^2 / 2) - G * M * m / (3R)Simplifying the equation:E = (1/4) * G * M * m / R - G * M * m / (3R)Combining terms:E = ( -3G * M * m + G * M * m ) / (12R)This simplifies to:E = 5GmM / 6RTherefore, the minimum energy required to launch the satellite (5GmM/6R) corresponds to option d in the question
Report an error
The more specific you are, the faster it gets fixed. A source beats an opinion.
Prefer email? support@testustad.com
About Gravitation
Newton’s law of universal gravitation relates the force between masses to their masses and separation, while gravitational field strength explains weight and orbital motion. The topic covers satellite velocity and period, escape ideas, weightlessness in free fall, and artificial gravity, which differs from ordinary gravitational attraction.
Practise Gravitation
265 free Gravitation MCQs from Physics, each with the correct answer and an explanation. Unlimited attempts, no account needed.
Discussion
Stuck on an option, or know a faster way to get there? Ask or explain it here.
No comments yet. Be the first to explain this one.
Exams that ask Physics questions like this
Physics is on 11 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.
More Gravitation questions
An object is taken from the earth to a planet where the value 'g' is 20 m/s2, if it is compared with value of 'g' earth, which of the following statement is true?
A man whose weight is 900 N on the earth's surface goes to a height, which is equal to twice the earth's radius, above the surface of the earth, then the new weight of the man will be:
The radius of Earth is 6 x 10^6m. What is the acceleration of a meteor, when it is 8 x 10^6m from the center of the Earth?
A space ship is coming towards earth from moon at a high speed. The speed of the air ship has to be reduced. A rocket is fired to lower its speed in a certain direction. The rocket should be fired: