Talha's Physics Academy
Gravitation - Critical Velocity
Q. Define Critical velocity and derive its formula. Find the value of proper speed.
Critical Velocity
"The constant horizontal velocity required to put a satellite into a stable circular orbit around the Earth is called Critical Velocity ($v_c$). It is also known as orbital speed or proper speed."
Figure: Satellite orbiting very close to the Earth's surface at altitude $h \ll R$.
Derivation
If a satellite is orbiting very close to the surface of the Earth, its altitude is extremely small compared to the radius of the Earth ($h \ll R$). In this case:
- The orbital radius is approximately equal to the radius of the Earth: $r = R + h \approx R$
- The gravitational acceleration at that altitude equals the surface gravity: $g_h = g$
- The orbital velocity becomes the critical velocity: $v = v_c$
From our general expression for orbital velocity:
$v = \sqrt{\frac{g R^2}{R + h}}$
Substituting $R + h \approx R$ into the equation:
$v_c = \sqrt{\frac{g R^2}{R}}$
Canceling one factor of $R$ yields the final formula for Critical Velocity:
$v_c = \sqrt{g R}$ --- (i)
Calculation of Proper Speed (Numerical Value)
Given standard values for a near-Earth orbit:
- Radius of Earth, $R = 6.38 \times 10^6 \, \text{m}$
- Gravitational acceleration, $g = 10 \, \text{m/s}^2$ (or $9.8 \, \text{m/s}^2$)
Substituting these values into equation (i):
$v_c = \sqrt{(10 \, \text{m/s}^2) \times (6.38 \times 10^6 \, \text{m})}$
$v_c = \sqrt{6.38 \times 10^7 \, \text{m}^2/\text{s}^2}$
$v_c = \sqrt{63,800,000} \approx 7987.5 \, \text{m/s}$
$v_c \approx 7.9 \, \text{km/s}$ (or $8 \, \text{km/s}$)
Physical Significance: As a satellite gets closer to the Earth, the gravitational pull gets stronger. Consequently, satellites in lower orbits must travel at a higher speed (approximately $7.9 \, \text{km/s}$) to maintain their orbit compared to those in farther orbits.
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