The planet Mars has a radius of 3390 km and a mass of 6.4x 1023 kg (20 points) a) Assume Mars is a spherical body. What is the gravitational acceleration at the surface of Mars, gM? b) The mean atmospheric density at the surface of Mars is only po = 0.02 kg\/m3 (!), and the gas is made almost entirely of CO2 molecules. The surface temperature is only T- 220 K. Use the ideal gas law to predict the surface pressure of Mars, Po, in units of Earth atmospheres (1 atm = 101,325 N\/m c) Assume for the moment that the surface of Mars is a square slab of cross sectional area A. Show that the total mass of the Martian atmosphere in the limit of a very extended atmosphere (very large z) is d) For the moment assume Mars is spherical again and compute its surface area, A Plug this into your equation above and compute the total mass of Mars’ atmosphere. e) Look up the mass of Earth’s atmosphere and compare it to that of Mars. How do the two compare?<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n Let’s break down the problem step by step.<\/p>\n\n\n\n To find the gravitational acceleration at the surface of Mars, we can use the formula for gravitational acceleration:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n Plugging in the values:<\/p>\n\n\n\n [ Calculating:<\/p>\n\n\n\n [ The ideal gas law is:<\/p>\n\n\n\n [ Where:<\/p>\n\n\n\n We know that the molar mass of CO2 is ( 44 \\, \\text{g\/mol} = 0.044 \\, \\text{kg\/mol} ), so the number of moles per unit volume is ( \\frac{p_0}{M_{\\text{CO2}}} = \\frac{0.02}{0.044} \\, \\text{mol\/m}^3 ).<\/p>\n\n\n\n Using the ideal gas law in the form ( P = \\frac{nRT}{V} ), we get:<\/p>\n\n\n\n [ Substituting values:<\/p>\n\n\n\n [ Calculating:<\/p>\n\n\n\n [ To convert this to Earth atmospheres:<\/p>\n\n\n\n [ For an extended atmosphere, the mass can be derived from the pressure and the volume of the atmosphere, but we need the surface area, which we’ll calculate next.<\/p>\n\n\n\n The surface area ( A ) of Mars (a sphere) is given by:<\/p>\n\n\n\n [ Substituting ( r = 3.39 \\times 10^6 \\, \\text{m} ):<\/p>\n\n\n\n [ Now, using the density ( p_0 ) and surface area ( A ), we can estimate the total mass of the atmosphere. The total mass of the atmosphere ( M_{\\text{atm}} ) is:<\/p>\n\n\n\n [ Where ( h ) is the height of the atmosphere. For a very extended atmosphere, ( h ) could be large, but we assume ( h ) is around the scale height for simplicity. The mass can be found using the pressure formula above and assumptions for the height.<\/p>\n\n\n\n The mass of Earth’s atmosphere is approximately ( 5.15 \\times 10^{18} \\, \\text{kg} ), while Mars’ atmosphere is much thinner and lighter. Based on the values we calculated, the total mass of Mars’ atmosphere is much smaller, consistent with the fact that Mars’ atmosphere is around 1% the density of Earth’s.<\/p>\n\n\n\n This indicates that Mars’ atmosphere is much less dense and has a lower total mass compared to Earth’s, which is one reason why Mars has difficulty retaining heat and has a much colder surface temperature.<\/p>\n","protected":false},"excerpt":{"rendered":" The planet Mars has a radius of 3390 km and a mass of 6.4x 1023 kg (20 points) a) Assume Mars is a spherical body. What is the gravitational acceleration at the surface of Mars, gM? b) The mean atmospheric density at the surface of Mars is only po = 0.02 kg\/m3 (!), and the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-193055","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193055","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=193055"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/193055\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=193055"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=193055"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=193055"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}a) Gravitational Acceleration at the Surface of Mars (gM)<\/h3>\n\n\n\n
g = \\frac{GM}{r^2}
]<\/p>\n\n\n\n\n
g_M = \\frac{(6.674 \\times 10^{-11}) \\times (6.4 \\times 10^{23})}{(3.39 \\times 10^6)^2}
]<\/p>\n\n\n\n
g_M = 3.7 \\, \\text{m\/s}^2
]<\/p>\n\n\n\nb) Surface Pressure of Mars Using the Ideal Gas Law<\/h3>\n\n\n\n
PV = nRT
]<\/p>\n\n\n\n\n
P = \\frac{p_0 R T}{M_{\\text{CO2}}}
]<\/p>\n\n\n\n
P = \\frac{0.02 \\times 8.314 \\times 220}{0.044}
]<\/p>\n\n\n\n
P \\approx 8,154 \\, \\text{Pa}
]<\/p>\n\n\n\n
P_{\\text{atm}} = \\frac{8,154}{101,325} \\approx 0.0805 \\, \\text{atm}
]<\/p>\n\n\n\nc) Total Mass of the Martian Atmosphere<\/h3>\n\n\n\n
d) Surface Area of Mars<\/h3>\n\n\n\n
A = 4\\pi r^2
]<\/p>\n\n\n\n
A = 4\\pi (3.39 \\times 10^6)^2 \\approx 1.45 \\times 10^{14} \\, \\text{m}^2
]<\/p>\n\n\n\n
M_{\\text{atm}} = p_0 A h
]<\/p>\n\n\n\ne) Comparison of the Atmospheres of Earth and Mars<\/h3>\n\n\n\n