Discussion
A satellite is moving around the earth in a circular orbit at a height of 2R, from the earth surface. R is the radius of the earth. The speed of satellite is :
Answer: C
The speed is \((gR/3)^{1/2}\).
At a height 2R above the surface the orbital radius is \(r = R + 2R = 3R\). Gravity provides the centripetal force, and \(GM = gR^{2}\) at the surface:
\[ \begin{aligned} \frac{mv^{2}}{r} &= \frac{GMm}{r^{2}} \\ v^{2} &= \frac{GM}{3R} = \frac{gR^{2}}{3R} = \frac{gR}{3} \\ v &= (gR/3)^{1/2} \end{aligned} \]
The option \((gR/2)^{1/2}\) takes the radius as 2R, forgetting to add the earth's radius to the height. The options gR/3 and gR/2 are values of \(v^{2}\), not of v, and do not have the dimensions of speed.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development