Question

An object is released from a distance ' \(r\) ' from the centre of the earth. The velocity of the object at the time of striking the earth will be ( \(R=\) radius of earth, \(r>R, g=\) acceleration due to gravity)

• A. \([2\mathrm{gR}(\mathrm{r}-\mathrm{R})/\mathrm{r}]^{\frac{1}{2}}\)
• B. \([gR(r-R)/r^{2}]^{\frac{1}{2}}\)
• C. \([2\mathrm{gR}(\mathrm{r}-\mathrm{R})/\mathrm{r}^{2}]^{\frac{1}{2}}\)
• D. \([gR^{2}(r-R)/r^{2}]^{\frac{1}{2}}\)

Hint:

Apply conservation of energy between distance r and the surface of the earth, using the relation between GM and g.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Gravitational potential energy at distance \(x\) from the centre is \(-\dfrac{GMm}{x}\). Total energy is conserved as the object falls from rest.

Step 2: Energy Balance
\(-\dfrac{GMm}{r}=-\dfrac{GMm}{R}+\dfrac{1}{2}mv^{2}\)
\(\dfrac{1}{2}v^{2}=GM\left(\dfrac{1}{R}-\dfrac{1}{r}\right)=\dfrac{GM(r-R)}{rR}\)

Step 3: Use \(GM=gR^{2}\)
\(v^{2}=\dfrac{2gR^{2}(r-R)}{rR}=\dfrac{2gR(r-R)}{r}\).

Step 4: Compare the options
\(v=\left[\dfrac{2gR(r-R)}{r}\right]^{1/2}\) is option (A).

Final Answer:
The speed on striking the earth is \(\left[2gR(r-R)/r\right]^{1/2}\). \[ \boxed{\text{(A) }[2\mathrm{gR}(\mathrm{r}-\mathrm{R})/\mathrm{r}]^{\frac{1}{2}}} \]