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The radius of Mercury (from the center to just above the atmosphere) is 2440 km (2440 × 103 m), and its mass is 3 × 1023 kg. An object is launched straight up from just above the atmosphere of Mercury.

a. What initial speed is needed so that when the object is far from Mercury its final speed is 2000 m/s?
b. What initial speed is needed so that when the object is far from Mercury its final speed is 0 m/s?

Respuesta :

akande212

Answer:

(a)4676m/s

(b) 4227m/s

Explanation:

Let the mass of mercury be Mm = 3×10²³ kg and the mass of the object be M

Also let the distance from the center of mercury be R1 = 2.240×10⁶m let the distance the object is launched to be R

From the energy conservation equation

K1 + U1 = K2 + U2

K1, K2 = initial and final kinetic energies.

U1, U2 = initial and final potential energies.

1/2MV1² + (-GMm×M/R1) = 1/2MV2² + (-GMm×M/R)

Assuming that the very far R approaches infinity then the second term to the right side of the equation above approaches zero

So dividing through by M since M is common to both sides we have that

1/2V1² + (-G×Mm/R1) = 1/2V2²

Rearranging,

V1² = 2×( 1/2V2²+ GMm/R1 )

Given V2 = 2000m/s and also the constant value of G = 6.67×10‐¹¹ N•m²/kg²

Substituting,

V1² =2×( 1/2×2000² + 6.67×10-¹¹ ×3×10²³ /(2.24×10⁶))

V1² = 21866071.43

V1 = 4676m/s

(b) this is still similar to the case described above but this time V2 = 0m/s

V1² = 2×( 1/2V2²+ GMm/R1 ) = 2×(1/2×0+ GMm/R1) = 2GMm/R1

V1² = 2×6.67×10-¹¹ × 3×10²³/(2.24×10⁶)

= 17866071.43

V1 = 4227m/s

Answer (a)

The initial speed is needed so that when the object is far from Mercury its final speed is 2000 m/s is 4676m/s.

Answer(b)

The initial speed is needed so that when the object is far from Mercury its final speed is 0 m/s is 4227m/s.

Speed

Answer (a)

The initial speed is needed so that when the object is far from Mercury its final speed is 2000 m/s is :

Let the mass of mercury be Mm = 3×10²³ kg .

Let the distance from the center of mercury = R1 = 2.240×10⁶m let

K1, K2 = initial and final kinetic energies.

U1, U2 = initial and final potential energies.

K1 + U1 = K2 + U2

1/2MV1² + (-GMm×M/R1) = 1/2MV2² + (-GMm×M/R)

1/2V1² + (-G×Mm/R1) = 1/2V2²

V1² = 2×( 1/2V2²+ GMm/R1 )

Given :

V2 = 2000m/s

G = 6.67×10‐¹¹ N•m²/kg²

Substituting the Values :

V1² =2×( 1/2×2000² + 6.67×10-¹¹ ×3×10²³ /(2.24×10⁶))

V1² = 21866071.43

V1 = 4676m/s

Answer (b)

The initial speed is needed so that when the object is far from Mercury its final speed is 0 m/s is:

V1² = 2×( 1/2V2²+ GMm/R1 ) = 2×(1/2×0+ GMm/R1) = 2GMm/R1

V1² = 2×6.67×10-¹¹ × 3×10²³/(2.24×10⁶)

V1 = 17866071.43

V1 = 4227m/s

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