Use Coupon: CART20 and get 20% off on all online Study Material

Total Price: Rs.

There are no items in this cart.
Continue Shopping
Grade: 11
Q. A body is projected upward from the Earth surface. It's velocity is half of the escape velocity. Find the total energy of it on the surface. If the maximum distance reached bu the body from earth - centre br r, find r.
one year ago

Answers : (1)

24739 Points
We know that the escape velocity of Earth = Ve = √[2gR]
       g = gravity due to Earth
      1/2 m (Ve)² - G Mm/R = 0   for the object to escape from Earth.
       (or,  KE + PE = 0)
When the body is projected from Earth's surface:
     Given  u = initial velocity = Ve /2 = √(gR/2)
     Initial  KE = 1/2 m u² = m g R/4 = G M m /(4R)
     Initial  PE = - GMm/R
     Total initial energy = - 3 GMm /(4R)
 When the body reaches a height h above surface of Earth and stops:
     v = 0. Final KE = 0.
     Final PE = - G M m/ (R+h)
From conservation of Energy:
           - GMm/(R+h) = - 3Gm/ (4R)
            3 (R+h) = 4 R
              h = R/3
one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Course Features

  • 731 Video Lectures
  • Revision Notes
  • Previous Year Papers
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Test paper with Video Solution

Course Features

  • 18 Video Lectures
  • Revision Notes
  • Test paper with Video Solution
  • Mind Map
  • Study Planner
  • NCERT Solutions
  • Discussion Forum
  • Previous Year Exam Questions

Ask Experts

Have any Question? Ask Experts

Post Question

Answer ‘n’ Earn
Attractive Gift
To Win!!! Click Here for details