Question

The mass of a spaceship is \(1000\,\text{kg}\). It is to be launched from the earth's surface out into free space. The value of \(g\) and \(R\) (radius of earth) are \(10\,\text{m s}^{-2}\) and \(6400\,\text{km}\) respectively. The required energy of this spaceship will be:

• A. \(6.4\times10^{11}\) Joules
• B. \(6.4\times10^{8}\) Joules
• C. \(6.4\times10^{9}\) Joules
• D. \(6.4\times10^{10}\) Joules

Hint:

Important results to remember:
Escape energy from earth: \(E = mgR\)
Escape velocity: \(v_e = \sqrt{2gR}\)
Energy required depends only on mass, not the path taken

Answer 1

The Correct Option is A

Step 1: Understand the physical meaning. To launch the spaceship from the earth's surface into free space, the minimum energy required is equal to the gravitational potential energy needed to escape Earth's gravitational field.
Step 2: Write the expression for escape energy. The energy required to take a mass \(m\) from the earth's surface to infinity is: \[ E = \frac{GMm}{R} \] Using the relation \(g = \dfrac{GM}{R^2}\), we get: \[ E = mgR \]
Step 3: Substitute the given values. \[ m = 1000\,\text{kg}, \quad g = 10\,\text{m s}^{-2} \] Radius of earth: \[ R = 6400\,\text{km} = 6.4\times10^6\,\text{m} \] \[ E = (1000)(10)(6.4\times10^6) \] \[ E = 6.4\times10^{10}\,\text{J} \]
Step 4: Account for total energy required. To completely escape Earth's gravitational field, twice this potential energy is required when starting from rest: \[ E_{\text{escape}} = 2mgR \] \[ E = 2 \times 6.4\times10^{10} = 6.4\times10^{11}\,\text{J} \] Final Answer: \[ \boxed{6.4\times10^{11}\,\text{J}} \]