The Question is from neet-physics gravitation satellites, and it was created by the author VENKATESWARARAO

Question

A satellite is in a circular orbit around Earth. If its orbital radius is doubled, how does its orbital speed change?

VENKATESWARARAO · Accepted Answer

## CORE INFORMATION
✓ **Answer**: [B]
## CONCEPT FRAMEWORK
	**Primary Concept**: Orbital Speed in Circular Motion
	**Related Topics**: Gravitational Force, Centripetal Force
	**Prerequisites**: Understanding of Newton's Law of Universal Gravitation and Centripetal Force
## SOLUTION PATHWAY
**Given Information**:
	* Satellite is in a circular orbit.
	* Orbital radius is doubled (r becomes 2r).
	* We need to find the change in orbital speed.
**Method & Steps**:
	1. Equate Gravitational Force to Centripetal Force
		Reasoning: In a circular orbit, the gravitational force provides the necessary centripetal force to keep the satellite in orbit.
		Formula/Rule: $F_g = G\frac{Mm}{r^2}$, $F_c = m\frac{v^2}{r}$
	2. Derive the Orbital Speed Formula
		Working: $G\frac{Mm}{r^2} = m\frac{v^2}{r}$. Simplifying, we get $v^2 = \frac{GM}{r}$, and $v = \sqrt{\frac{GM}{r}}$.
		Key Point: Note that the orbital speed, v, is inversely proportional to the square root of the radius, r.
	3. Analyze the change in speed.
		Result: If r becomes 2r, the new speed, $v' = \sqrt{\frac{GM}{2r}} = \frac{1}{\sqrt{2}}\sqrt{\frac{GM}{r}} = \frac{v}{\sqrt{2}}$. Therefore, the speed decreases by a factor of $\sqrt{2}$.
		Verification: The new speed, v', is the original speed v divided by the square root of 2, meaning it decreases by a factor of $\sqrt{2}$ 
## OPTION ANALYSIS
[A] :x: [Incorrect. The speed decreases, not increases.]
[B] ✓: [Correct. The speed decreases by a factor of $\sqrt{2}$ when the radius is doubled.]
[C] :x: [Incorrect. The speed changes with the orbital radius.]
[D] :x: [Incorrect. The speed decreases, not increases.]
## LEARNING AIDS
	**Common Mistakes**:
		1.[Error 1]: Forgetting the square root relationship between orbital speed and radius. [Review the derivation of the orbital speed formula to avoid this error]
		2.[Error 2]: Confusing the relationship between radius and speed. [Remember that as radius increases, speed decreases in an orbital situation.]
	**Quick Tips**:
		1.[Helpful shortcut]: Remember the formula for orbital speed $v = \sqrt{\frac{GM}{r}}$ and how it relates to radius.
		2.[Memory aid]: Think of a planet further from the sun moving slower in its orbit. The same principle applies to satellites orbiting Earth.

VENKATESWARARAO · Answer

Increases by a factor of 2

VENKATESWARARAO · Answer

The correct Option is "Decreases by a factor of $\sqrt{2}$"

VENKATESWARARAO · Answer

Remains the same

VENKATESWARARAO · Answer

Increases by a factor of $\sqrt{2}$

NEET Physics Gravitation Satellites MCQs

Select Topic

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

Question Tags

NEET Subjects

Our Bots

Company