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soldier1979 [14.2K]

3 years ago

9

Which celestial body would have the strongest gravitational pull on a satellite orbiting 100 km above its surface?

2 answers:

marshall27 [118]3 years ago

4 0

:Sample Response: Jupiter would have the strongest gravitational pull on a satellite orbiting above its surface because gravity is directly proportional to mass and Jupiter is the most massive planet.

Explanation:

8090 [49]3 years ago

3 0

According to the Law of Universal Gravitation, the gravitational force is directly proportional to the mass, and inversely proportional to the distance. In this problem, let's assume the celestial bodies to be restricted to the planets and the Sun. Since the distance is specified, the other factor would be the mass. Among all the celestial bodies, the Sun is the most massive. So, the Sun would cause the strongest gravitational pull to the satellite.

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2 years ago

Read 2 more answers

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KengaRu [80]

Answer:

(A). The speed of the ions is $1.2\times10^{6}\ m/s$

(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$

Explanation:

Given that,

Electric field = 60000 N/C

Magnetic field = 0.0500 T

(A). We need to calculate the velocity

For no deflection

$F_{E}=F_{B}$

$Eq=Bqv$

$v = \dfrac{E}{B}$

$v=\dfrac{60000}{0.0500}$

$v=1.2\times10^{6}\ m/s$

(B). We need to calculate the radius

Using magnetic force balance by centripetal force

$Bqv=\dfrac{mv^2}{r}$

$r=\dfrac{mv^2}{Bqv}$

Put the value into the formula

$r=\dfrac{1.16\times10^{-26}\times(1.2\times10^{6})^2}{0.0500\times1.6\times10^{-19}}$

$r=2.0\times10^{6}\ m$

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(B). The radius of curvature of a singly charged lithium ion is $2.0\times10^{6}\ m$

6 0

2 years ago