 Click to Chat

1800-1023-196

+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
`        (a) Find the radius of the circular orbit of a satellite moving with an angular speed equal to the angular speed of speed of earth’s rotation. (b) If the satellite is directly above the north pole at some instant, find the time it takes to come over the equatorial plane. Mass of the earth = 6 × 1024 kg.`
5 years ago

```							Sol. Angular speed f earth & the satellite will be same
2π/T base e = 2π/T base s
Or 1/24 * 3600 = 1/2π√(R + h)^3/gR^2 or 12 l 3600 = 3.14 √(R + h)^3/gR^2
Or √(R + h)^3/gR^2 = (12 * 3600)^2/(3.14)^2 or (6400 + h)^3 + 10^9/9.8 * (6400)^2 * 10^6 = (12 * 3600)^2/(3.14)^2
Or (6400 + h)^3 * 10^9/6272 * 10^9 = 432 * 10^4
Or (6400 + h)^3 = 6272 * 432 * 10^4
Or 6400 + h = (6272 * 432 * 10^4)^1/3
Or h = (6272 * 432 * 10^4)^1/3 – 6400
= 42300 cm.
b) Time taken from north pole to equator = (1/2) t
= (1/2) * 6.28 √(43200 + 6400)^3/10 * (6400)^2 * 10^6 = 3.14 √(497)^3 * 10^6/(64)^2 * 10^11
= 3.14 √497 * 497 * 497/64 * 64 * 10^5 = 6 hour.

```
5 years ago
Think You Can Provide A Better Answer ?

## Other Related Questions on Mechanics

View all Questions »  ### 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