Exams › JEE Main › Physics

A particle is launched vertically upward from Earth’s surface with speed equal to three-fourths of the escape speed. The maximum distance of the particle from the Earth’s center is (take Earth’s radius as R):

1. 10R/9
2. 16R/7
3. 9R/8
4. 10R/3

Correct answer: 16R/7

Solution

v0 = (3/4)ve, ve^2 = 2GM/R. Energy conservation: (1/2)v0^2 - GM/R = -GM/r_max gives (9/16)(GM/R) - GM/R = -GM/r_max -> -7/16 GM/R = -GM/r_max -> r_max = 16R/7.

Related JEE Main Physics questions

• A satellite of mass m moves in a circular orbit around the Earth at an altitude x above the Earth's surface. If g denotes the acceleration due to gravity at the Earth's surface, then the satellite's orbital speed is
• If Earth’s radius became 1% smaller while its mass stayed unchanged, what would happen to the acceleration due to gravity at the surface?
• The Moon takes about 29 days to complete one revolution around the Earth. If the Moon’s mass were doubled while every other factor stayed the same, its orbital period would be approximately
• Let the Earth have mean radius R, angular velocity ω about its own axis, and surface gravitational acceleration g. The orbital radius of a geostationary satellite is:
• Assume the Earth is a uniform sphere. Scientist A descends into a mine, while scientist B rises in a balloon. As they move, the gravitational field experienced by A keeps decreasing and the field experienced by B also keeps decreasing. Which statement best describes their rates of decrease?
• A ball is released from a satellite orbiting the Earth at an altitude of 120 km. What will happen to the ball?

⚔️ Practice JEE Main Physics free + battle 1v1 →