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If the mass of the earth is M, its radiu...

If the mass of the earth is M, its radius is R and gravitational constant G, then the speed required to be given to a projectile so that it does not return to earth is

A

`sqrt((GM)/R)`

B

`sqrt((2GM)/2)`

C

`sqrt((GR)/M)`

D

`sqrt(GRM)`

Text Solution

AI Generated Solution

The correct Answer is:
To find the speed required to be given to a projectile so that it does not return to Earth, we can use the principles of gravitational potential energy and kinetic energy. Here’s a step-by-step solution: ### Step 1: Understand the Problem We need to determine the speed (v) that a projectile must have to escape Earth's gravitational pull. This speed is often referred to as the escape velocity. ### Step 2: Write Down the Given Data - Mass of the Earth: \( M \) - Radius of the Earth: \( R \) - Gravitational constant: \( G \) ### Step 3: Use the Conservation of Energy The total mechanical energy at the surface of the Earth must equal the total mechanical energy when the projectile is at an infinite distance (where potential energy is zero). The initial total energy (E_initial) is the sum of kinetic energy (KE) and gravitational potential energy (PE): \[ E_{\text{initial}} = KE + PE = \frac{1}{2} mv^2 - \frac{GMm}{R} \] Where: - \( m \) is the mass of the projectile - \( v \) is the speed of the projectile The final total energy (E_final) when the projectile is far away (at infinity) is: \[ E_{\text{final}} = 0 \] Thus, we set up the equation: \[ \frac{1}{2} mv^2 - \frac{GMm}{R} = 0 \] ### Step 4: Rearrange the Equation From the equation above, we can isolate the kinetic energy term: \[ \frac{1}{2} mv^2 = \frac{GMm}{R} \] ### Step 5: Cancel the Mass of the Projectile Since \( m \) appears on both sides of the equation, we can cancel it out (assuming \( m \neq 0 \)): \[ \frac{1}{2} v^2 = \frac{GM}{R} \] ### Step 6: Solve for v Now, multiply both sides by 2 to isolate \( v^2 \): \[ v^2 = \frac{2GM}{R} \] Taking the square root of both sides gives us the escape velocity: \[ v = \sqrt{\frac{2GM}{R}} \] ### Final Answer The speed required to be given to a projectile so that it does not return to Earth is: \[ v = \sqrt{\frac{2GM}{R}} \] ---
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