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2 years ago

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A satellite encircles Mars at a distance above its surface equal to 3 times the radius of Mars. If g m is the acceleration due t

o gravity at the surface of Mars, what is the acceleration due to gravity at the location of the satellite?

1 answer:

Sav [38]2 years ago

6 0

Acceleration due to gravity at the surface of Mars is given by the formula

$g_m = \frac{GM}{R^2}$

here

M = mass of Mars

R = radius of Mars

Now if we take a point which is at height 3 times the radius of Mars from its surface

So the acceleration due to gravity is given as

$g = \frac{GM}{(3R+R)^2}$

now using this

$g = \frac{GM}{16R^2}$

$g = \frac{g_m}{16}$

so here the gravity is 16 times lesser than gravity at surface

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2 years ago

Which of the following expressions will have units of kg⋅m/s2? Select all that apply, where x is position, v is velocity, m is m

netineya [11]

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Explanation:

In the image attached with this answer are shown the given options from which only one is correct.

The correct expression is:

$m \frac{d}{dt}v_{(t)}$

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The other expressions are incorrect, let’s prove it:

$\frac{m}{2} \frac{d}{dx}{(v_{(x)})}^{2}=\frac{m}{2} 2v_{(x)}^{2-1}=mv_{(x)}$ This result has units of $kg\frac{m}{s}$

$m\frac{d}{dt}a_{(t)}=ma_{(t)}^{1-1}=m$ This result has units of $kg$

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$m\frac{d}{dt}x_{(t)}=mx_{(t)}^{1-1}=m$ This result has units of $kg$

$m\frac{d}{dt}v_{(t)}=mv_{(t)}^{1-1}=m$ This result has units of $kg$

$\frac{m}{2}\int {(v_{(t)})}^{2} dt= \frac{m}{2} \frac{{(v_{(t)})}^{2+1}}{2+1}+C=\frac{m}{6} {(v_{(t)})}^{3}+C$ This result has units of $kg \frac{m^{3}}{s^{3}}$ and $C$ is a constant

$m\int a_{(t)} dt= \frac{m {a_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{4}}$ and $C$ is a constant

$\frac{m}{2} \frac{d}{dt}{(v_{(x)})}^{2}=0$ because $v_{(x)}$ is a constant in this derivation respect to $t$

$m\int v_{(t)} dt= \frac{m {v_{(t)}}^{2}}{2}+C$ This result has units of $kg \frac{m^{2}}{s^{2}}$ and $C$ is a constant

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