Question:
An object is thrown directly away from the surface of the earth with an initial speed \(V\). The object reaches upto a height of
\[
\frac{4}{5}R_E
\]
from earth’s surface, where \(R_E\) is radius of the earth. If the escape velocity of the object is \(V_E\), then the value of
\[
\frac{V}{V_E}
\]
is
An object is thrown directly away from the surface of the earth with an initial speed \(V\). The object reaches upto a height of
\[
\frac{4}{5}R_E
\]
from earth’s surface, where \(R_E\) is radius of the earth. If the escape velocity of the object is \(V_E\), then the value of
\[
\frac{V}{V_E}
\]
is
Show Hint
For variable gravitational potential problems, always use conservation of mechanical energy with
\[
U=-\frac{GMm}{r}
\]
instead of constant-\(g\) equations.
Updated On: Jun 26, 2026
- \(\frac{4}{3}\)
- \(\frac{3}{4}\)
- \(\frac{2}{3}\)
- \(\frac{4}{5}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use conservation of mechanical energy.
Let the mass of the object be \(m\).
Initially, the object is at earth’s surface: \[ r_i=R_E \] Final maximum distance from earth’s center is \[ r_f=R_E+\frac{4}{5}R_E \] \[ r_f=\frac{9}{5}R_E \] At maximum height, velocity becomes zero.
Step 2: Write the energy equation.
Initial total energy: \[ \frac{1}{2}mV^2-\frac{GMm}{R_E} \] Final total energy: \[ -\frac{GMm}{r_f} \] Thus, \[ \frac{1}{2}mV^2-\frac{GMm}{R_E} = -\frac{GMm}{\frac{9}{5}R_E} \] \[ \frac{1}{2}mV^2 = \frac{GMm}{R_E}-\frac{5GMm}{9R_E} \] \[ \frac{1}{2}mV^2 = \frac{4GMm}{9R_E} \] Hence, \[ V^2=\frac{8GM}{9R_E} \]
Step 3: Write the escape velocity expression.
Escape velocity is \[ V_E=\sqrt{\frac{2GM}{R_E}} \] Thus, \[ V_E^2=\frac{2GM}{R_E} \]
Step 4: Find the ratio.
\[ \left(\frac{V}{V_E}\right)^2 = \frac{\frac{8GM}{9R_E}}{\frac{2GM}{R_E}} \] \[ = \frac{8}{18} \] \[ = \frac{4}{9} \] Therefore, \[ \frac{V}{V_E}=\frac{2}{3} \]
Step 5: Final conclusion.
Hence, \[ \boxed{\frac{2}{3}} \]
Let the mass of the object be \(m\).
Initially, the object is at earth’s surface: \[ r_i=R_E \] Final maximum distance from earth’s center is \[ r_f=R_E+\frac{4}{5}R_E \] \[ r_f=\frac{9}{5}R_E \] At maximum height, velocity becomes zero.
Step 2: Write the energy equation.
Initial total energy: \[ \frac{1}{2}mV^2-\frac{GMm}{R_E} \] Final total energy: \[ -\frac{GMm}{r_f} \] Thus, \[ \frac{1}{2}mV^2-\frac{GMm}{R_E} = -\frac{GMm}{\frac{9}{5}R_E} \] \[ \frac{1}{2}mV^2 = \frac{GMm}{R_E}-\frac{5GMm}{9R_E} \] \[ \frac{1}{2}mV^2 = \frac{4GMm}{9R_E} \] Hence, \[ V^2=\frac{8GM}{9R_E} \]
Step 3: Write the escape velocity expression.
Escape velocity is \[ V_E=\sqrt{\frac{2GM}{R_E}} \] Thus, \[ V_E^2=\frac{2GM}{R_E} \]
Step 4: Find the ratio.
\[ \left(\frac{V}{V_E}\right)^2 = \frac{\frac{8GM}{9R_E}}{\frac{2GM}{R_E}} \] \[ = \frac{8}{18} \] \[ = \frac{4}{9} \] Therefore, \[ \frac{V}{V_E}=\frac{2}{3} \]
Step 5: Final conclusion.
Hence, \[ \boxed{\frac{2}{3}} \]
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