A satellite of mass m is launched vertically upwards with an initial speed `sqrt((GM)/®)` from the surface of the earth. After it reaches height R (R = radius of the earth), it ejects a rocket of mass `m/10` in a direction opposite to the initial direction of the satellite, so that subsequently the satellite escapes to infinity. The minimum kinetic energy of the rocket at ejection needed is (G is the gravitational constant, M is the mass of the earth):
A
`81/20 .(GMm)/(R )`
B
`21/20. (GMm)/(R )`
C
`81/10. (GMm)/(R )`
D
`21/40. (GMm)/(R )`
Text Solution
Verified by Experts
The correct Answer is:
A
Energy conservation: `- (GMm)/(R)+ 1/2 m (sqrt((GM)/(R))) =- (GMm)/(2R) + 1/2 m v _(1) ^(2)` `v _(1) =0` Momentum conservation:
A satellite of mass m is launched vertically upwards with an initial speed U from the surface of the Earth. After it reaches height 2 R (R = radius of Earth) it ejects a rocket of mass (m)/(5) So, that subsequently the satellite moves in a circular orbit. The kinetic energy of rocket is (M is mass of Earth)
A satellite of mass m is launched vertically upwards with initial kinetic energy K from the Earth’s surface. After it reaches height R (R = radius of earth). It ejects a rocket of mass m/5 so that subsequently the satellite moves in a circular orbit. The kinetic energy of rocket is [G is gravitational constant, M is the mass of the earth, take (GMm)/(R)=K ]
A body is projected vertically upwards with a speed of sqrt((GM)/R) ( M is mass and R is radius of earth ) . The body will attain a height of
A satellite of mass 'M' is projected radially from surface of earth with speed 'u'. When is reaches a height equal to radius of earth, it ejects a rocket of mass ( M)/(10 ) and it itself starts orbiting the earth in circular path of radius 2R, find the kinetic energy of rocket.
A satellite of mass m_0 is revolving around the earth in circular orbit at a height of 3R from surface of the earth the areal velocity of satellite is (where R is radius of earth and M_0 is the mass of earth)
A satellite of mass m revolves around the earth of radius R at a hight x from its surface. If g is the acceleration due to gravity on the surface of the earth, the orbital speed of the satellite is
What is the energy required to launch a m kg satellite from earth's surface in a circular orbit at an altitude of 7R ? (R = radius of the earth)
A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:
A satellite of mass m is revolving around the Earth at a height R above the surface of the Earth. If g is the gravitational intensity at the Earth’s surface and R is the radius of the Earth, then the kinetic energy of the satellite will be:
A rocket of mass M is launched vertically from the surface of the earth with an initial speed V= sqrt((gR)/(2)) . Assuming the radius of the earth to be R and negligible air resistance, the maximum height attained by the rocket above the surface of the earth is (R)/(X) where X is:
VMC MODULES ENGLISH-JEE MAIN REVISON TEST-23-PHYSICS (SECTION 2)
