Click to Chat

1800-2000-838

+91-120-4616500

CART 0

• 0

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
`        the height above the surface of earth where g decreases to 9% of its maximum value on earth surface is.(radius of earth 6400 km)`
7 years ago

147 Points
```										Dear anand s
g=G Me/(Re +h)2
wher h is hight from earth surface
value of g on earth  surface is for h=0
g1=G Me/(Re )2

so we need to find g such that
g=.09 g1
G Me/(Re +h)2   =.09G Me/(Re )2
or Re =.3 (Re+h)
h=7Re/3

Please feel free to post as many doubts on our discussion forum as you can. If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.  All the best.  Regards, Askiitians Experts Badiuddin
```
7 years ago
Think You Can Provide A Better Answer ?

Other Related Questions on General Physics

View all Questions »
• Complete JEE Main/Advanced Course and Test Series
• OFFERED PRICE: Rs. 15,900
• View Details
• Kinematics & Rotational Motion
• OFFERED PRICE: Rs. 636
• View Details