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3 years ago

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A 6500 kg satellite orbits a planet. If its distance to the center of the planet is 9e6 m and its speed is 6.7e3 m/s, what is th

e gravitational force between the satellite and the planet

2 answers:

otez555 [7]3 years ago

6 0

Complete Question

The complete question is shown on the first uploaded image

Answer:

Option D is the Answer

Explanation:

The formula for the velocity of the satellite is

$v =\sqrt{\frac{GM}{R} }$

R is the distance to the center o the planet = $9*10^6m$

G is the gravitational constant which is equal to $= 6.67*10^{-11} \frac{Nm^2}{kg^2}$

M is the mass of the planet

Making M the subject of the formula we have

$M = \frac{(v^2)(R)}{G}$

$M = \frac{(6.7*10^3)^2(9*10^6)}{6.67*10^{-11}} = 60.57*10^{23}Kg$

The formula for the gravitational force between two object

$F = \frac{GMm}{R^2}$

Where

$= \frac{6.67*10^{-11}(6.057*10^{24})(6500)}{(9*10^6)^2}$

$F = 32,419.90 N$

$= 3.24*10^{4} N$

Ira Lisetskai [31]3 years ago

3 0

Answer:

the gravitational force between the satellite and the planet is 32.4 * 10³N

Explanation:

It is given that,

Mass of the satellite, m = 6500 kg

Speed of the satellite, v = 6.7 × 10³ m/s

distance to the center of the planet = 9 × 10⁶m

Let F is the magnitude of the gravitational force exerted on the satellite by the planet. The centripetal force is equal to the gravitational force. It is equal to :

$F = \frac{mv^2}{R}$

$F = \frac{6500 \times (6.7 * 10^3)}{9 * 10^6}$

= 32.4 * 10³N

the gravitational force between the satellite and the planet is 32.4 * 10³N

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Answer:

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