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One of your summer lunar space camp activities is to launch a 1090 kg rocket from the surface of the Moon. You are a serious space camper and you launch a serious rocket: it reaches an altitude of 211 km . What gain Δ???? in gravitational potential energy does the launch accomplish? The mass and radius of the Moon are 7.36×1022 kg and 1740 km, respectively.

User Parnell
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1 Answer

3 votes

Answer:

ΔP.E = 6.48 x 10⁸ J

Step-by-step explanation:

First we need to calculate the acceleration due to gravity on the surface of moon:

g = GM/R²

where,

g = acceleration due to gravity on the surface of moon = ?

G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²

M = Mass of moon = 7.36 x 10²² kg

R = Radius of Moon = 1740 km = 1.74 x 10⁶ m

Therefore,

g = (6.67 x 10⁻¹¹ N.m²/kg²)(7.36 x 10²² kg)/(1.74 x 10⁶ m)²

g = 2.82 m/s²

now the change in gravitational potential energy of rocket is calculated by:

ΔP.E = mgΔh

where,

ΔP.E = Change in Gravitational Potential Energy = ?

m = mass of rocket = 1090 kg

Δh = altitude = 211 km = 2.11 x 10⁵ m

Therefore,

ΔP.E = (1090 kg)(2.82 m/s²)(2.11 x 10⁵ m)

ΔP.E = 6.48 x 10⁸ J

User Vinit Sarvade
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