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Georgia [21]
2 years ago
12

NASA sends an unmanned lander to test conditions on the surface of Mars. What is the magnitude of the gravitational force acting

on the lander on the surface of Mars? DATA: Mass of Mars = 6.42·1023 kg Radius of Mars = 3397 km Mass of lander = 138.1 kg
Physics
1 answer:
kozerog [31]2 years ago
4 0

Answer:

The magnitude of the gravitational force acting on the lander on the surface of Mars is 512.46 N.

Explanation:

The Universal law of gravity is define as:

F = G \frac{M.m}{R^{2}}  (1)

Where F is the gravitational force, G is gravitational constant, M is the mass of Mars, m is the mass of the lander and R is the radius of Mars.

Before replacing the values in equation 1 it is necessary to express the radius of mars in terms of meters:

R = 3397 kmx\frac{1000 m}{1 km} ⇒ R = 3397000 m

Finally, equation 1 can be used:

F = (6.67x10^{-11} N.m^{2}/Kg^{2}) \frac{(6.42x10^{23} kg)(138.1 kg)}{(3397000 m)^{2}}

F = 512.46 N

Hence, the magnitude of the gravitational force acting on the lander on the surface of Mars is 512.46 N.

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The question is incomplete. Here is the complete question.

Cars A nad B are racing each other along the same straight road in the following manner: Car A has a head start and is a distance D_{A} beyond the starting line at t = 0. The starting line is at x = 0. Car A travels at a constant speed v_{A}. Car B starts at the starting line but has a better engine than Car A and thus Car B travels at a constant speed v_{B}, which is greater than v_{A}.

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Part B: How far from Car B's starting line will the cars be when Car B passes Car A? Express your answer in terms of known quantities.

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Explanation: First, let's write an equation of motion for each car.

Both cars travels with constant speed. So, they are an uniform rectilinear motion and its position equation is of the form:

x=x_{0}+vt

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The equation will be:

x_{A}=D_{A}+v_{A}t

Car B started at the starting line. So, its equation is

x_{B}=v_{B}t

Part A: When they meet, both car are at "the same position":

D_{A}+v_{A}t=v_{B}t

v_{B}t-v_{A}t=D_{A}

t(v_{B}-v_{A})=D_{A}

t=\frac{D_{A}}{v_{B}-v_{A}}

Car B meet with Car A after t=\frac{D_{A}}{v_{B}-v_{A}} units of time.

Part B: With the meeting time, we can determine the position they will be:

x_{B}=v_{B}(\frac{D_{A}}{v_{B}-v_{A}} )

x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}}

Since Car B started at the starting line, the distance Car B will be when it passes Car A is x_{B}=\frac{v_{B}D_{A}}{v_{B}-v_{A}} units of distance.

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