What is the 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 of 2r?

India – the minimum energy required to launch a satellite of mass m the minimum energy required to launch a satellite of mass m– We are going to start the discussion about THE MINIMUM ENERGY REQUIRED TO LAUNCH A SATELLITE OF MASS M as per our readers’ demands and comments. If you want to know about this India topic, continue reading and learn more. • • • • • • • • About the minimum energy required to launch a satellite of mass m What is the 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 of 2R. A 6R5GmM B 3R2GmM C 2RGmM D 3RGmM Medium Solution Verified by Toppr Corect option is A) Given that, Mass of satellite =m Mass of planet =M Radius =R Altitude h=2R Now,. People Also Read: What is The minimum energy required to launch a m kg satellite from The minimum energy required to launch amkg satellite from earth's surface in a circular orbit at an altitude of 2R where R is the radius of earth, will be: A 3mgR B 65mgR C 2mgR D 51mgR Medium Solution Verified by Toppr Corect option is B) The kinetic energy at altitude 2R is = 6RGMm The gravitational potential energy at altitude 2R is = 3R–GMm. What is the 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 of 2R. A 3R2GmM B 2RGmM C 3RGmM D 6R5GmM Medium Solution Verified by Toppr Corect option is D) Given that, Mass of satellite =m Mass of planet =...

Solution: Given that, Mass of satellite = m Mass of planet = M Radius = R Altitude h = 2 R Now, The gravitational potential energy P.E = r − G m ​ Potential energy at altitude = 3 R G m M ​ Orbital velocity v 0 ​ = R + h G m M ​ Now, the total energy is E f ​ = 2 1 ​ m v 0 2 ​ − 3 R G m M ​ E f ​ = 2 1 ​ 3 R G m M ​ − 3 R GM m ​ E f ​ = 3 R G m M ​ [ 2 1 ​ − 1 ] E f ​ = 6 R − G m M ​ Now, E i ​ = E f ​ Now, the minimum required energy K ⋅ E = R G mm ​ − 6 R G m M ​ K ⋅ E = 6 R 5 G m M ​ Hence, the minimum required energy is 6 R 5 G m M ​

This problem has been solved! You'll get a detailed solution from a subject matter expert that helps you learn core concepts. See Answer See Answer See Answer done loading Question:What is the minimum energy required to launch a satellite of mass m kg from the earth’s surface of radius R in a circular orbit at an altitude of 2R ?

Exp. (a) From conservation of energy, Total energy at the planet = Total energy at altitude − R GM m ​ + ( K E ) s u r f a ce ​ = − 3 R GM m ​ + 2 1 ​ m v A 2 ​ … ( i ) In the orbital of planet, the necessary centripetal force is obtained by gravitational force. ∴ R + 2 R m v A 2 ​ ​ = ( R + 2 R ) 2 GM m ​ ⇒ 2 v A ​ ​ = 3 R GM ​ From Eqs. (i) and (ii), we get ( KE ) s u r f a ce ​ = 6 5 ​ R GMm ​ Views: 5,572 R is made of two semi-circular parts (see figure). The two parts are held together by a ring made of a metal strip of cross-sectional area S and length L . L is slightly less than 2 π R. To fit the ring on the wheel, it is heated so that its temperature rises by Δ T and it just steps over the wheel. As it cools down to surrounding temperature, it presses the semi-circular parts together. If the coefficient of linear expansion of the metal is α and its Young's modulus is Y, the force that one part of the wheel applies on the other part is (a) 2 π S Y α Δ T (b) S Y α Δ T (c) π S Y α Δ T (d) 2 S Y α Δ T Views: 5,458 h' above earth surface (where, radius of earth = 6.4 × 1 0 3 km ) is E 1 ​ and kinetic energy required for the satellite to be in a circular orbit at this height is E 2 ​ . The value of h for which E 1 ​ and E 2 ​ are equal is (a) 3.2 × 1 0 3 km (b) 1.28 × 1 0 4 km (c) 6.4 × 1 0 3 km (d) 1.6 × 1 0 3 km Views: 6,080 g E ​ and g M ​ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment ...

Concept: • The total mechanical energy of a satellite is the sum of its kinetic energy (always positive) and potential energy (may be negative). • At infinity, the gravitational potential energy of the satellite is zero. • As the Earth-satellite system is a bound system, the total energy of the satellite is negative. • Generally, the total energy of the planet andsatellite system is negative. This means the satellite cannot escape from the earth’s gravity. ​ The energy of the satellite is calculated as = \(\frac\)

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