Question

The critical velocity of a satellite revolving around the Earth is 10 km/s at a height where g_h = 8 m/s². Calculate the height of the satellite from the surface of the Earth. (R = 6.4 × 10⁶ m)

Answer 1

Given: v_c = 10 km/s = 10 × 10³ m/s,

g_h = 8 m/s³, R = 6.4 × 10⁶ m

To find: Height of the satellite (h)

Formula: v_c = \(\sqrt{g_h(R+h)}\)

Calculation: From formula,

10 × 10³ = \(\sqrt{8\times(R+h)}\)

Squaring both the sides, we get,

100 × 10⁶ = 8(R + h)

∴ 8(R + h) = 100 × 10⁶

∴ R + h = \(\frac{100}{8}\) × 10⁶

∴ h = 12.5 × 10⁶ – R

= 12.5 × 10⁶ – 6.4 × 10⁶

= 6.1 × 10⁶m

∴ h = 6100 km

The height of the satellite from the surface of the Earth is 6100 km.