An object starts vertically with speed Vo. Prove that its speed V at height h is given by V 2 o -V 2 = (2gh/1+h/R) where R is the radius of the earth. The earth is a solid sphere of radius 6.4 x 10 6 m and g = 9.8 ms -2 at its surface. Find the vertical speed with which an object should be thrown so as to reach up to height from the surface of the earth. Briefly explain the concept of escape velocity.

Physics
An object starts vertically with speed Vo. Prove that its speed V at height h is given by V 2 o -V 2 = (2gh/1+h/R) where R is the radius of the earth. The earth is a solid sphere of radius 6.4 x 10 6 m and g = 9.8 ms -2 at its surface. Find the vertical speed with which an object should be thrown so as to reach up to height from the surface of the earth. Briefly explain the concept of escape velocity.

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Answer

v2=v02−2ghv^2 = v_0^2 - 2 g h

Part (a)

Step 1: Apply conservation of mechanical energy. Assume potential energy is zero at the starting point.

Initial energy: kinetic energy + potential energy = 12mv02+0\frac{1}{2} m v_0^2 + 0

Final energy at height hh: 12mv2+mgh\frac{1}{2} m v^2 + m g h

12mv02=12mv2+mgh\frac{1}{2} m v_0^2 = \frac{1}{2} m v^2 + m g h

Step 2: Divide both sides by mm:

12v02=12v2+gh\frac{1}{2} v_0^2 = \frac{1}{2} v^2 + g h

Step 3: Multiply both sides by 22:

v02=v2+2ghv_0^2 = v^2 + 2 g h

Step 4: Rearrange for v2v^2:

v2=v02−2ghv^2 = v_0^2 - 2 g h

v^2 = v_0^2 - 2 g h

Part (b)

The object reaches maximum height h=Rh = R from the surface, so at r=2Rr = 2R from Earth's center, v=0v = 0. Use conservation of energy with gravitational potential energy −GMmr-\frac{G M m}{r}.

Step 1: Initial energy at r=Rr = R: 12mv02−GMmR\frac{1}{2} m v_0^2 - \frac{G M m}{R}

Final energy at r=2Rr = 2R: 0−GMm2R0 - \frac{G M m}{2 R}

12mv02−GMmR=−GMm2R\frac{1}{2} m v_0^2 - \frac{G M m}{R} = -\frac{G M m}{2 R}

Step 2: Divide by mm:

12v02−GMR=−GM2R\frac{1}{2} v_0^2 - \frac{G M}{R} = -\frac{G M}{2 R}

Step 3: Add GMR\frac{G M}{R} to both sides:

12v02=GMR−GM2R=GM2R\frac{1}{2} v_0^2 = \frac{G M}{R} - \frac{G M}{2 R} = \frac{G M}{2 R}

Step 4: Multiply by 22:

v02=GMRv_0^2 = \frac{G M}{R}

Step 5: Use surface gravity g=GMR2g = \frac{G M}{R^2}, so GM=gR2G M = g R^2:

v02=gR2R=gRv_0^2 = \frac{g R^2}{R} = g R v0=gRv_0 = \sqrt{g R}

Step 6: Substitute g=9.8 m/s2g = 9.8 \, \mathrm{m/s^2}, R=6.4×106 mR = 6.4 \times 10^6 \, \mathrm{m}:

gR=9.8×6.4×106g R = 9.8 \times 6.4 \times 10^6

9.8×6.4=62.729.8 \times 6.4 = 62.72

gR=6.272×107 m2/s2g R = 6.272 \times 10^7 \, \mathrm{m^2/s^2}

Step 7: v0=6.272×107 m/sv_0 = \sqrt{6.272 \times 10^7} \, \mathrm{m/s}

6.272×107=6.272×107≈2.504×3162.3≈7920 m/s\sqrt{6.272 \times 10^7} = \sqrt{6.272} \times \sqrt{10^7} \approx 2.504 \times 3162.3 \approx 7920 \, \mathrm{m/s}

\mathbf{7920 m/s}

Part (c)

Escape velocity is the minimum speed an object must be given from Earth's surface to reach infinity (r→∞r \to \infty) with zero final speed, overcoming Earth's gravity completely.

Using energy conservation:

12mvesc2−GMmR=0\frac{1}{2} m v_esc^2 - \frac{G M m}{R} = 0 vesc=2GMR=2gRv_esc = \sqrt{\frac{2 G M}{R}} = \sqrt{2 g R}
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Quick Answer

Part (a) Step 1: Apply conservation of mechanical energy. Assume potential energy is zero at the starting point.

An object starts vertically with speed Vo. Prove that its speed V at height h is given by V 2 o -V 2 = (2gh/1+h/R) where R is the radius of the earth. The earth is a solid sphere of radius 6.4 x 10 6 m and g = 9.8 ms -2 at its surface. Find the vertical speed with which an object should be thrown so as to reach up to height from the surface of the earth. Briefly explain the concept of escape velocity.
Physics

This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.

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Part (a) Step 1: Apply conservation of mechanical energy. Assume potential energy is zero at the starting point. Initial energy: kinetic energy + potential energy = (1)/(2) m v_0^2 + 0 Final energy at height h: (1)/(2) m v^2 + m g h (1)/(2) m v_0^2 = (1)/(2) m v^2 + m g h Step 2: Divide both sides by m: (1)/(2) v_0^2 = (1)/(2) v^2 + g h Step 3: Multiply both sides by 2: v_0^2 = v^2 + 2 g h Step 4: Rearrange for v^2: v^2 = v_0^2 - 2 g h v^2 = v_0^2 - 2 g h Part (b) The object reaches maximum height h = R from the surface, so at r = 2R from Earth's center, v = 0. Use conservation of energy with gravitational potential energy -(G M m)/(r). Step 1: Initial energy at r = R: (1)/(2) m v_0^2 - (G M m)/(R) Final energy at r = 2R: 0 - (G M m)/(2 R) (1)/(2) m v_0^2 - (G M m)/(R) = -(G M m)/(2 R) Step 2: Divide by m: (1)/(2) v_0^2 - (G M)/(R) = -(G M)/(2 R) Step 3: Add (G M)/(R) to both sides: (1)/(2) v_0^2 = (G M)/(R) - (G M)/(2 R) = (G M)/(2 R) Step 4: Multiply by 2: v_0^2 = (G M)/(R) Step 5: Use surface gravity g = (G M)/(R^2), so G M = g R^2: v_0^2 = (g R^2)/(R) = g R v_0 = sqrt(g R) Step 6: Substitute g = 9.8 \, m/s^2, R = 6.4 × 10^6 \, m: g R = 9.8 × 6.4 × 10^6 9.8 × 6.4 = 62.72 g R = 6.272 × 10^7 \, m^2/s^2 Step 7: v_0 = sqrt(6.272 × 10^7) \, m/s sqrt(6.272 × 10^7) = sqrt(6.272) × sqrt(10^7) ≈ 2.504 × 3162.3 ≈ 7920 \, m/s 7920 m/s Part (c) Escape velocity is the minimum speed an object must be given from Earth's surface to reach infinity (r ) with zero final speed, overcoming Earth's gravity completely. Using energy conservation: (1)/(2) m v_esc^2 - (G M m)/(R) = 0 v_esc = sqrt((2 G M)/(R)) = sqrt(2 g R)