MY CART (5)

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

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping
Menu
Grade: 9
        
Calculate the position of the planet 10^5 kg which is placed between the planets of 10^20kg and 10^24 kg which are placed at 6000km apart hence it will get zero gravitational force.(I have solved the question,I only want the working and Correct and complete answer) 
2 months ago

Answers : (1)

Vikas TU
8721 Points
							
Hiii 
Gravitational force = GM1*M2/d2
Let us assume distance of 10^5 Kg particles from 10^20 Kg be x 
So gravitation force between them = G *10^25 /x2
Gravitational force between 10^5 and 10^24 kg mass will be = G* 10^29 /(6000-x)2
It is getting a zero gravitational force so, G*10^25 /x2 = G* 10^29 /(6000-x)2
(6000 – x)2  = 10^4 * x2
solve for x and apply quadratic equation formula , 
You will get X = 59.40 Km 
2 months ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies


Course Features

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


Course Features

  • 110 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
Vouchers
To Win!!! Click Here for details